题解

26 条题解

  • 2
    @ 2018-11-02 11:13:10

    有人说枚举会超时
    可是我就是枚举做的

    #include<bits/stdc++.h>
    using namespace std;
    int bc[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
    bool check(int x)
    {
        int tmp=x,n=0;
        while(tmp>=10)
        {
            n=n*10+tmp%10;
            tmp/=10;
        }
        n=n*10+tmp;
        //cout<<n<<" "<<x<<":\n";
        if(n==x) return true;
        return false;
    }
    int main()
    {
        int a,b,cnt=0;scanf("%d%d",&a,&b);
        for(int i=a;i<=b;i++)
        {
            bool flag=0;//是不是闰年 
            int day=i%100;
            int mon=(i/100)%100;
            int year=i/10000;
            if(year%4==0)
            {
                if(year%100==0)
                {
                    if(year%400==0)
                     flag=1;
                }
                else flag=1;
            }
            if(flag)
            {
                if(mon==2)
                {
                    if(day>29) continue;
                }
                else if(day > bc[mon]) continue;
            }
            else if(day > bc[mon]) continue;
            if(mon > 12) continue;
            if(check(i))
            {
                //printf("%d\n",i);
                cnt++;  
            }
        }
        printf("%d",cnt);
        return 0;
    }
    
  • 1
    @ 2022-01-12 18:02:47

    啊确实我就是枚举做的

    var
    yy,mm,dd,sum,yydl,mmdl,dddl,l,i,dx,k:longint;
    s1,s2,sl,yz,mz,dz:string;
    zw,hw:array[1..8]of char;
    ding,ohh:boolean;
    begin
    readln(s1);
    val(s1[1..4],yy,l);val(s1[5..6],mm,l);val(s1[7..8],dd,l);
    readln(s2);
    val(s2[1..4],yydl,l);val(s2[5..6],mmdl,l);val(s2[7..8],dddl,l);
    ding:=false;
    while ding=false do
    begin
    str(yy,yz);
    for i:=1 to 4 do zw[i]:=yz[i];
    if(mm<10)then
    begin
    zw[5]:='0';
    str(mm,sl);
    zw[6]:=sl[1];
    end else begin
    str(mm,mz);
    for i:=5 to 6 do zw[i]:=mz[i-4]
    end;
    if(dd<10)then
    begin
    zw[7]:='0';
    str(dd,sl);
    zw[8]:=sl[1];
    end else begin
    str(dd,dz);
    for i:=7 to 8 do zw[i]:=dz[i-6];
    end;
    for i:=1 to 8 do hw[i]:=zw[9-i];
    ohh:=true;k:=0;
    repeat
    inc(k);
    if zw[k]<>hw[k]then ohh:=false;
    until(ohh=false)or(k=8);
    if(k=8)and(ohh=true)then inc(sum);
    if(yy=yydl)and(mm=mmdl)and(dd=dddl)then ding:=true
    else begin
    if(mm=1)or(mm=3)or(mm=5)or(mm=8)or(mm=7)or(mm=10)or(mm=12)then dx:=31
    else if(mm=4)or(mm=6)or(mm=9)or(mm=11)then dx:=30
    else begin
    if(yy mod 4=0)and(yy mod 100<>0)or(yy mod 400=0)then dx:=29
    else dx:=28;
    end;
    inc(dd);
    if dd>dx then
    begin
    dd:=1;
    inc(mm);
    end;
    if mm>12then
    begin
    mm:=1;
    inc(yy);
    end;
    end;
    end;
    write(sum);readln;readln;
    end.

  • 1
    @ 2021-08-30 09:04:37
    #include <bits/stdc++.h>
    using namespace std;
    
    int main()
    {
        int date1,date2,year1,year2,fan_year,aim_date,tmp_year,month,day,cnt=0;
        int days[13] = {0,31,29,31,30,31,30,31,31,30,31,30,31};
        cin>>date1>>date2;
        year1=date1/10000;
        year2=date2/10000;
        for(int year=year1; year<=year2; year++){
            tmp_year=year;
            fan_year=0;
            for(int i=1; i<=4; i++){
                fan_year=tmp_year%10+fan_year*10;
                tmp_year/=10;
            }
            month=fan_year/100;
            day=fan_year%100;
            aim_date=year*10000+month*100+day;
            if(month<=12 && month>=1 && day<=days[month] && aim_date>=date1 && aim_date<=date2)
                cnt++;
        }
        cout<<cnt;
        return 0;
    }
    
  • 1
    @ 2017-10-07 18:45:08

    这题其实不用枚举,时间消耗比较多。。(分秒必争呀)
    其实先把年份枚举,再反推末四位,看符不符合。
    注意2月29日的回文日期为9220年02月29日,特判一下就可以了,其他的2月只要超过28就return false
    感觉数据还是水了点,其实我的代码有些漏洞呀。。等于首尾年份是应该再看看是否在那个范围里(还好数据水)
    代码:

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    bool dao(int f)
    {
        int aa=f%10;
        f=f/10;
        int bb=f%10;
        f=f/10;
        int cc=f%10,dd=f/10;
        int ab=aa*10+bb,dc=cc*10+dd;
        if(ab==2&&dc==29) return true;
        if(ab>12||ab<1) return false;
        if((dc>31)||((ab==4||ab==6||ab==9||ab==11)&&dc==31)||(ab==2&&dc>=29)) return false;
        return true;
    }
    int main()
    {
        int a,b,c,d,e=0;
        cin>>a>>b;
        c=a/10000;
        d=b/10000;
        for(int i=c;i<=d;++i)
        {
            if(dao(i)==true) e++;
        }
        cout<<e;
        return 0;
    }
    
  • 1
    @ 2017-09-17 10:52:21

    寀鸡

    • @ 2017-09-17 10:52:55

      大哥威武!!!

    • @ 2018-09-02 19:10:53

      hhhhh

      
      
  • 0
    @ 2020-04-13 15:16:36
    /*
    根本不需要枚举, 注意每年中最多有一天是回文就行
    */
    #include <iostream>
    #include <algorithm>
    #include <string>
    using namespace std;
    
    bool Bigyear(int year)
    {
        if(year % 4 != 0 || (year % 100 == 0 && year % 400 != 0))
            return false;
        return true;
    }
    
    bool Date(string date)
    {
        int year, month, day;
        year = stoi(date.substr(0, 4));
        month = stoi(date.substr(4, 2));
        day = stoi(date.substr(6, 2));
        if(month <= 0 || month >= 13 || day <= 0 || day >= 32)
            return false;
        if(month == 4 || month == 6 || month == 9 || month == 11)
        {
            if(day == 31)
                return false;
        }
        else if(month == 2)
        {
            if(Bigyear(year))
            {
                if(day >= 30)
                    return false;
            }
            else
                if(day >= 29)
                    return false;
        }
        return true;
    }
    
    int judge(string date1, string date2, string date)
    {
        if(stoi(date) >= stoi(date1) && stoi(date) <= stoi(date2) && Date(date))
            return 1;
        return 0;
    }
    
    int Ans(string date1, string date2)
    {
        int ans = 0;
        string s1, s2;
        int k = stoi(date1.substr(0, 4));
        int M = stoi(date2.substr(0, 4));
        while (k <= M)
        {
            s2 = s1 = to_string(k);
            reverse(s2.begin(), s2.end());
            s1 += s2;
            ans += judge(date1, date2, s1);
            k++;
        }
        return ans;
    }
    
    int main()
    {
        string date1, date2;
        cin >> date1 >> date2;
    
        cout << Ans(date1, date2) << endl;
    
        return 0;
    }
    
    
  • 0
    @ 2018-01-23 01:29:14

    吐槽一下Vijos的C编译器居然不认识gets()&&itoa();
    May the father of understanding guide U;

  • 0
    @ 2017-02-15 13:33:06
    #include <cstdio>
    int m[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
    inline bool check(int date) {
        int t[9];
        t[0] = 0;
        while (date) {
            t[++t[0]] = date%10;
            date /= 10;
        }
        for (int i = 1;i <= 4;i++)
            if (t[i] != t[8-i+1]) return false;
        return true;
    }
    inline int next(int i) {
        int year,month,day;
        day = (i%10)+((i/10%10)*10);
        i /= 100;
        month = (i%10)+((i/10%10)*10);
        i /= 100;
        year = (i%10)+((i/10%10)*10)+((i/100%10)*100)+i/1000*1000;
        if ((!(year%4) && year%100) || !(year%400)) m[2] = 29;
        day++;
        if (day == m[month]+1) {
            day = 1;
            month++;
        }
        if (month == 13) {
            month = 1;
            year++;
        }
        return day+month*100+year*10000;
    }
    int main() {
        //freopen("date.in","r",stdin);
        //freopen("date.out","w",stdout);
        int date1,date2,ans = 0;
        scanf("%d%d",&date1,&date2);
        for (int i = date1;i <= date2;i = next(i))
            if (check(i)) ans++;
        printf("%d",ans);
        return 0;
    }
    
  • 0
    @ 2017-02-15 09:15:56

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #1: Accepted, time = 15 ms, mem = 728 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #3: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 732 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 728 KiB, score = 10
    测试数据 #8: Accepted, time = 93 ms, mem = 732 KiB, score = 10
    测试数据 #9: Accepted, time = 156 ms, mem = 736 KiB, score = 10
    Accepted, time = 279 ms, mem = 736 KiB, score = 100

    这题太缩水了。。。

  • 0
    @ 2017-02-15 09:07:08

    大家想一想,这道题千万不能用枚举,不要相信任何枚举内容

  • 0
    @ 2017-02-15 09:06:54
    #include <bits/stdc++.h>
    using namespace std;
    
    bool hw(int a)
    {
        int x[11];
        for(int i=0;i<8;a/=10,i++) {
            x[i]=a%10;
        }
        for(int i=0,j=7;i<4;i++,j--) {
            if(x[i]!=x[j]) return 0;
        }
        return 1;
    }
    int main()
    {
        int i,j,k,l=0,n,m;
        cin>>n>>m;
        int x=n;
        for(;x<=m;) {
            bool run=false;
            if(x/10000%400==0||(x/10000%100!=0&&x/10000%4==0)) run=true;
            short flag;
            k=x/100%100;
            if(k==1||k==3||k==5||k==7||k==8||k==10||k==12) flag=2;
            else if(k==2) {
                if(run) flag=0;
                else {
                    flag=-1;
                }
            }
            else flag=1;
            if(flag==2&&x%100==32) x+=69;
            else if(flag==1&&x%100==31) x+=70;
            else if(flag==0&&x%100==30) x+=71;
            else if(flag==-1&&x%100==29) x+=72;
            k=x/100%100;
            if(k==13) x=x+8800;
            if(hw(x)) l++;
            x++;
        }
        cout<<l<<endl;
        return 0;
    } 
    
  • 0
    @ 2017-02-15 09:06:42
    #include <bits/stdc++.h>
    using namespace std;
    
    bool hw(int a)
    {
        int x[11];
        for(int i=0;i<8;a/=10,i++) {
            x[i]=a%10;
        }
        for(int i=0,j=7;i<4;i++,j--) {
            if(x[i]!=x[j]) return 0;
        }
        return 1;
    }
    int main()
    {
        int i,j,k,l=0,n,m;
        cin>>n>>m;
        int x=n;
        for(;x<=m;) {
            bool run=false;
            if(x/10000%400==0||(x/10000%100!=0&&x/10000%4==0)) run=true;
            short flag;
            k=x/100%100;
            if(k==1||k==3||k==5||k==7||k==8||k==10||k==12) flag=2;
            else if(k==2) {
                if(run) flag=0;
                else {
                    flag=-1;
                }
            }
            else flag=1;
            if(flag==2&&x%100==32) x+=69;
            else if(flag==1&&x%100==31) x+=70;
            else if(flag==0&&x%100==30) x+=71;
            else if(flag==-1&&x%100==29) x+=72;
            k=x/100%100;
            if(k==13) x=x+8800;
            if(hw(x)) l++;
            x++;
        }
        cout<<l<<endl;
        return 0;
    } 
    ```c++
    
  • -1
    @ 2017-08-25 00:24:51

    so water

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    int mon[13]={-1,31,28,31,30,31,30,31,31,30,31,30,31};
    int main()
    {
        int d1,d2,f=0;
        scanf("%d\n%d",&d1,&d2);
        int y1=d1/10000;
        int y2=d2/10000;
        int sum=0;
        for(int i=y1;i<=y2;i++)
        {
            int d=i*10000+i%10*1000+i%100/10*100+i%1000/100*10+i%10000/1000;
            int m=d%10000/100;
            int day=d%100;
            if(m>=1&&m<=12)
            {
                if(day>=1&&day<=mon[m])
                    sum++;
            }
            if(i==9220)
                sum++;
        }
        printf("%d",sum);
        return 0;
    }
    
  • -1
    @ 2017-07-27 23:02:05

    超长代码
    const c:array[1..12] of integer=(31,29,31,30,31,30,31,31,30,31,30,31);
    var n,i,s,k:longint;
    a,b:array[1..3] of integer;
    s1,s2:string;
    procedure re;
    var i,sx:integer;
    begin
    sx:=1000;
    for i:=1 to 8 do
    begin
    if i<5 then begin
    a[1]:=a[1]+(ord(s1[i])-48)*sx;
    sx:=sx div 10;
    end;
    if (i<7)and(i>4) then begin
    a[2]:=a[2]+(ord(s1[i])-48)*sx;
    sx:=sx div 10
    end;
    if (i<9)and(i>6) then begin
    a[3]:=a[3]+(ord(s1[i])-48)*sx;
    sx:=sx div 10;
    end;
    if (i=4)or(i=6) then sx:=10;
    end;
    end;
    procedure rea;
    var i,sx:integer;
    begin
    sx:=1000;
    for i:=1 to 8 do
    begin
    if i<5 then begin
    b[1]:=b[1]+(ord(s2[i])-48)*sx;
    sx:=sx div 10;
    end;
    if (i<7)and(i>4) then begin
    b[2]:=b[2]+(ord(s2[i])-48)*sx;
    sx:=sx div 10;
    end;
    if (i<9)and(i>6) then begin
    b[3]:=b[3]+(ord(s2[i])-48)*sx;
    sx:=sx div 10;
    end;
    if (i=4)or(i=6) then sx:=10;
    end;
    end;
    procedure pp(x:integer);
    begin
    if ((x mod 4=0)and(x mod 100<>0))or(x mod 400=0) then c[2]:=29
    else c[2]:=28;
    end;
    function check(y,z:string):boolean;
    begin
    if (y[1]=z[4])and(y[2]=z[3])and(y[3]=z[2])and(y[4]=z[1]) then exit(true)
    else exit(false);
    end;
    begin
    readln(s1);
    readln(s2);
    if s1=s2 then
    begin
    s2:=copy(s1,5,4);
    delete(s1,5,4);
    if check(s1,s2) then writeln(1)
    else writeln(0);
    halt;
    end;
    fillchar(a,sizeof(a),0);re;
    fillchar(b,sizeof(b),0);rea;
    s:=0;
    pp(a[1]);
    while (a[1]<>b[1])or(a[2]<>b[2])or(a[3]<>b[2]) do
    begin
    k:=a[2]*100+a[3];
    s1:='';s2:='';str(a[1],s1);str(k,s2);
    if a[2]<10 then insert('0',s2,1);
    if check(s1,s2) then inc(s);
    inc(a[3]);
    if a[3]>c[a[2]] then
    begin
    a[3]:=1;
    inc(a[2]);
    end;
    if a[2]>12 then
    begin
    a[2]:=1;
    inc(a[1]);
    pp(a[1]);
    end;
    end;
    writeln(s);
    end.

  • -1
    @ 2017-07-25 13:34:28

    完全不用枚举
    根据年份构造回文串判断是否合法即可
    强悍的O(1)
    cpp
    #include<bits/stdc++.h>
    using namespace std;
    int mon[13]={-1,31,28,31,30,31,30,31,31,30,31,30,31};
    int main()
    {
    int d1,d2,f=0;
    cin>>d1>>d2;
    int y1=d1/10000;
    int y2=d2/10000;
    int sum=0;
    for(int i=y1;i<=y2;i++)
    {
    int d=i*10000+i%10*1000+i%100/10*100+i%1000/100*10+i%10000/1000;
    int m=d%10000/100;
    int day=d%100;
    //cout<<d<<" "<<i<<"/"<<m<<"/"<<day<<endl;
    if(m>=1&&m<=12)
    {
    if(day>=1&&day<=mon[m])sum++;//cout<<d<<" "<<i<<"/"<<m<<"/"<<day<<endl;
    }
    if(i==9220)sum++;
    }
    cout<<sum<<endl;
    return 0;
    }

  • -1
    @ 2017-07-15 11:05:42

    帮我看看错哪了
    var
    s1,s2:string;
    s,i,j,y1,y2,m,d,x,y:longint;
    f:byte;
    begin
    readln(s1);readln(s2);
    val(copy(s1,1,4),y1);val(copy(s2,1,4),y2);
    for i:=y1 to y2 do begin
    f:=0;
    m:=i mod 10*10+i mod 100 div 10;
    d:=i div 100 mod 10*10+i div 1000;
    if (m>=1)and(m<=12) then f:=1;
    if (f=1) then begin
    if (m=1)or(m=3)or(m=5)or(m=7)or(m=8)or(m=10)or(m=12) then begin
    if (d>=1)and(d<=31) then f:=2;
    end
    else begin
    if (m<>2) then if (d>=1)and(d<=30) then f:=2
    else begin
    if (i mod 4=0)and(i mod 100<>0)or(i mod 400=0) then begin
    if (d>=1)and(d<=29) then f:=2;
    end
    else if (d>=1)and(d<=28) then f:=2;
    end;
    end;
    end;
    if i=y1 then begin
    val(copy(s1,5,2),x);val(copy(s1,7,2),y);
    if (f=2)and(m>=x)and(d>=y)then inc(s);
    end;
    if (i=y2)and(i<>y1) then begin
    val(copy(s2,5,2),x);val(copy(s2,7,2),y);
    if (f=2)and(m<=x)and(d<=y)then inc(s);
    end;
    if (i>y1)and(i<y2)then if f=2 then inc(s);
    end;
    writeln(s);
    end.

  • -1
    @ 2017-07-15 11:04:06

    var
    s1,s2:string;
    s,i,j,y1,y2,m,d,x,y:longint;
    f:byte;
    begin
    readln(s1);readln(s2);
    val(copy(s1,1,4),y1);val(copy(s2,1,4),y2);
    for i:=y1 to y2 do begin
    f:=0;
    m:=i mod 10*10+i mod 100 div 10;
    d:=i div 100 mod 10*10+i div 1000;
    if (m>=1)and(m<=12) then f:=1;
    if (f=1) then begin
    if (m=1)or(m=3)or(m=5)or(m=7)or(m=8)or(m=10)or(m=12) then begin
    if (d>=1)and(d<=31) then f:=2;
    end
    else begin
    if (m<>2) then if (d>=1)and(d<=30) then f:=2
    else begin
    if (i mod 4=0)and(i mod 100<>0)or(i mod 400=0) then begin
    if (d>=1)and(d<=29) then f:=2;
    end
    else if (d>=1)and(d<=28) then f:=2;
    end;
    end;
    end;
    if i=y1 then begin
    val(copy(s1,5,2),x);val(copy(s1,7,2),y);
    if (f=2)and(m>=x)and(d>=y)then inc(s);
    end;
    if (i=y2)and(i<>y1) then begin
    val(copy(s2,5,2),x);val(copy(s2,7,2),y);
    if (f=2)and(m<=x)and(d<=y)then inc(s);
    end;
    if (i>y1)and(i<y2)then if f=2 then inc(s);
    end;
    writeln(s);
    end.

  • -1
    @ 2017-03-22 14:26:15

    主要考判断语句,从开始年份到目标年份枚举,每年月份枚举,注意月份的天数有的31有的30,还要判断是不是为闰年or平年,如此便穷举出所有有效日期,随后判断计数输出即可。
    ```pascal
    var
    d1,d2,sy,ey,i,j,k,s,ans:longint;

    function pan(s:longint):boolean;
    var
    i:longint;
    st:string;
    begin
    str(s,st);
    for i:= 1 to 4 do
    begin
    if st[i]<>st[8-i+1] then exit(false);
    end;
    exit(true);
    end;
    begin
    readln(d1,d2);
    sy:=d1 div 10000;
    ey:=d2 div 10000;
    for i:= sy to ey do
    begin
    for j:= 1 to 12 do
    begin
    if (j=1) or (j=3) or(j=5) or (j=7) or (j=8) or (j=10) or (j=12) then
    begin
    for k:= 1 to 31 do
    begin
    s:=i*10000+j*100+k;
    if (s>=d1) and (s<=d2) then
    if pan(s) then inc(ans);
    end;
    end
    else if (j=2) then
    begin
    if ( (i mod 4 =0) and (i mod 100 <>0 )) or (i mod 400=0) then
    begin
    for k:= 1 to 29 do
    begin
    s:=i*10000+j*100+k;
    if (s>=d1) and (s<=d2) then
    if pan(s) then inc(ans);
    end;
    end
    else
    begin
    for k:= 1 to 28 do
    begin
    s:=i*10000+j*100+k;
    if (s>=d1) and (s<=d2) then
    if pan(s) then inc(ans);
    end;
    end;
    end
    else
    begin
    for k:= 1 to 30 do
    begin
    s:=i*10000+j*100+k;
    if (s>=d1) and (s<=d2) then
    if pan(s) then inc(ans);
    end;
    end;
    end;
    end;
    writeln(ans);
    end.
    ```

    • @ 2017-03-22 14:26:46
      var
      d1,d2,sy,ey,i,j,k,s,ans:longint;
      
      function pan(s:longint):boolean;
      var
      i:longint;
      st:string;
      begin
          str(s,st);
          for i:= 1 to 4 do
          begin
              if st[i]<>st[8-i+1] then exit(false); 
          end;
          exit(true);
      end;
      begin
          readln(d1,d2);
          sy:=d1 div 10000;
          ey:=d2 div 10000;
          for i:= sy to ey do
          begin
              for j:= 1 to 12 do
              begin
                  if (j=1) or (j=3) or(j=5) or (j=7) or (j=8) or (j=10) or (j=12) then
                  begin
                      for k:= 1 to 31 do
                      begin
                          s:=i*10000+j*100+k; 
                          if (s>=d1) and (s<=d2) then 
                          if pan(s) then inc(ans);
                      end;
                  end
                  else if (j=2) then 
                  begin
                      if ( (i mod 4 =0) and (i mod 100 <>0 )) or (i mod 400=0) then 
                      begin
                          for k:= 1 to 29 do
                          begin
                              s:=i*10000+j*100+k;
                              if (s>=d1) and (s<=d2) then 
                              if pan(s) then inc(ans);
                          end;
                      end
                      else
                      begin
                          for k:= 1 to 28 do
                          begin
                              s:=i*10000+j*100+k;
                              if (s>=d1) and (s<=d2) then 
                              if pan(s) then inc(ans);
                          end;
                      end;
                  end
                  else 
                  begin
                      for k:= 1 to 30 do
                      begin
                          s:=i*10000+j*100+k;
                          if (s>=d1) and (s<=d2) then 
                          if pan(s) then inc(ans);
                      end; 
                  end;
              end;
          end;
          writeln(ans);
      end.
      
    • @ 2017-07-18 17:16:14

      强悍的枚举
      居然没有超时
      我记得我考试的时候用枚举结果一个个tle

    • @ 2018-09-02 19:12:40

      @xieshihua: NA

  • -1
    @ 2017-02-18 20:27:49

    #include <iostream>
    #include<algorithm>
    using namespace std;
    int Sovle(int a[],int x,int n,int R)
    {
    int i=x,j;
    while(a[i]-R<a[x]&&i<n)
    {
    i++;
    }
    if(i==n&&a[i]-R<a[x])
    return 1;
    i--;
    for(j=i;j<n;j++)
    if(a[j]>a[i]+R)
    break;
    return 1+Sovle(a,j,n,R);
    }
    int main ()
    {
    int R,x[10000],N;
    cin>>R>>N;
    for(int i=0;i<N;i++)
    cin>>x[i];
    sort(x,x+N);
    cout<<Sovle(x,0,N,R);
    return 0;
    }

  • -1
    @ 2017-02-18 17:30:13

    #include <iostream>
    #include <string>
    using namespace std;
    int get(string s,bool x)
    {
    if(x==true)
    return (s[0]-48)*1000+(s[1]-48)*100+(s[2]-48)*10+(s[3]-48);
    else
    return (s[7]-48)*1000+(s[6]-48)*100+(s[5]-48)*10+(s[4]-48);
    }
    void change(int x,int a[])
    {
    int i,xx=x;
    for(i=0;i<4;i++)
    a[i]=xx%10,xx/=10;
    }
    bool judge(int a[],int year)
    {
    int month=a[0]*10+a[1],day=a[2]*10+a[3];
    if(month==1||month==3||month==5||month==7||month==8||month==10||month==12)
    {
    if(day<=31)
    {
    return true;
    }
    }
    if(month==4||month==6||month==9||month==11)
    {
    if(day<=30)
    {
    return true;
    }
    }
    if(month==2)
    {
    if(year%4==0&&year%100!=0||year%400==0)
    {
    if(day<=29)
    {
    return true;
    }
    }
    else
    {
    if(day<=28)
    {
    return true;
    }
    }
    }
    return false;
    }
    int main ()
    {
    string start,end;
    cin>>start>>end;
    int s=get(start,true),e=get(end,true),a[4],sum=0;
    if(get(start,true)<get(start,false))
    s++;
    if(get(end,true)<get(end,false))
    e--;
    while(s<=e)
    {
    change(s,a);
    if(judge(a,s)==true)
    sum++;
    s++;
    }
    cout<<sum;
    return 0;
    }

信息

ID
2010
难度
4
分类
(无)
标签
递交数
1302
已通过
323
通过率
25%
被复制
18
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