82 条题解

  • 2
    @ 2018-03-24 20:40:48
    #include<iostream>
    #include<cstdio>
    #include<vector>
    #include<map>
    #define FOR(i,x,y) for(i=x;i<=y;++i)
    #define maxa 1000+10
    using namespace std;
    int a[10];
    int main()
    {
        string s;
        cin>>s;
        int len = s.length();
        if((s[s.length()-1]-'0')%2==0)
            a[1] = 1;
        else
          a[1] = 0;
          int sum =0,i;
        for(i=0;i<len;++i)
            sum+=s[i]-'0';
        if(sum%3==0)
            a[2] = 1;
        else
            a[2] = 0;
       if(((s[len-2]-'0')*10+s[len-1]-'0')%4==0)
            a[3] = 1;
        else
          a[3] = 0;
       if(((s[len-3]-'0')*100+(s[len-2]-'0')*10+s[len-1]-'0')%8==0)
            a[4] = 1;
        else
          a[4] = 0;
    
        if(s[len-1]=='1'||s[len-1]=='4'||s[len-1]=='6'||s[len-1]=='9'||s[len-1]=='0')
            a[5] = 1;
        else
              a[5] = 0;
          for(i=1;i<=5;++i)
            cout<<a[i]<<endl;
        return 0;
    }
    
    
    
  • 1
    @ 2019-01-31 22:48:10

    #include <stdio.h>
    #include <math.h>
    #include <string.h>
    int main ()
    {
    char c[100000];
    scanf("%s",c);
    int len=strlen(c);
    int a=c[len-1]-'0';
    int b=(c[len-2]-'0')*10+a;
    int d=(c[len-3]-'0')*10+b;
    int e=0,i;
    for(i=0;i<len;i++)
    e+=c[i];
    if (a%2==0)
    printf("1\n");
    else
    printf("0\n");
    if (e%3==0)
    printf("1\n");
    else
    printf("0\n");
    if (b%4==0)
    printf("1\n");
    else
    printf("0\n");
    if (d%8==0)
    printf("1\n");
    else
    printf("0\n");
    if (a!=2&&a!=3&&a!=7&&a!=8)
    printf("1\n");
    else
    printf("0\n");
    }

  • 0
    @ 2016-12-20 17:43:48

    我操。。。。。数组一定要开大。。。rte 两次 百度发现开大数组 然后ac

  • 0
    @ 2016-08-16 20:53:57

    #include<iostream>
    #include<cstring>
    using namespace std;
    char a[1000];
    int ein=1,zwei=0;
    long int gw=0;
    int main()
    {
    cin>>a;
    int len=strlen(a);
    if((a[len-1]-'0')%2==0)
    cout<<ein;
    else
    cout<<zwei;
    cout<<endl;
    for(int i=0;i<len;i++)
    gw+=a[i]-'0';
    if(gw%3==0)
    cout<<ein;
    else
    cout<<zwei;
    cout<<endl;
    if(((a[len-2]-'0')*10+a[len-1]-'0')%4==0)
    cout<<ein;
    else
    cout<<zwei;
    cout<<endl;
    if(((a[len-3]-'0')*100+(a[len-2]-'0')*10+a[len-1]-'0')%8==0)
    cout<<ein;
    else
    cout<<zwei;
    cout<<endl;
    if(a[len-1]=='1'||a[len-1]=='4'||a[len-1]=='6'||a[len-1]=='9'||a[len-1]=='0')
    cout<<ein;
    else
    cout<<zwei;
    return 0;
    }

  • 0
    @ 2016-06-20 16:29:34

    **
    末位判断2的倍数
    各个位之和判断3的倍数
    末两位判断4的倍数
    末三位判断5的倍数
    末位014569判断完全平方数
    **

    
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    string s;
    int x;
    int main(){
        getline(cin,s);
        for(unsigned i = 0;i < s.length();i++) x += s[i] - '0';
        if(s[s.length() - 1] & 1) printf("0\n");
        else printf("1\n");
        if(x % 3 == 0)printf("1\n");
        else printf("0\n");
        if((s[s.length() - 1] - '0' + (s[s.length() - 2] - '0') * 10) % 4 == 0) printf("1\n");
        else printf("0\n");
        if((s[s.length() - 1] - '0' + (s[s.length() - 2] - '0') * 10 + (s[s.length() - 3] - '0') * 100) % 8 == 0) printf("1\n");
        else printf("0\n");
        if(s[s.length() - 1] == '0' || s[s.length() - 1] == '1' || s[s.length() - 1] == '4' || s[s.length() - 1] == '5' || s[s.length() - 1] == '6' || s[s.length() - 1] == '9') printf("1\n");
        else printf("0\n");
        return 0;
    }
    
    
    
  • 0
    @ 2015-09-30 14:51:45

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  • 0
    @ 2015-08-24 17:51:32

    #include<iostream>
    #include<string>
    #include<set>
    using namespace std;

    int main() {
    string s; cin>>s;
    cout<<!(int(s[s.size()-1]-'0')%2)<<endl;
    int sum=0;
    for (int i=0;i<s.size();++i) sum+=s[i]-'0';
    cout<<!(sum%3)<<endl;
    sum=0;
    for (int i=(s.size()-2<0?0:s.size()-2);i<s.size();++i) sum=sum*10+(s[i]-'0');
    cout<<(!(sum%4))<<endl;
    for (int i=(s.size()-3<0?0:s.size()-3);i<s.size();++i) sum=sum*10+(s[i]-'0');
    cout<<(!(sum%8))<<endl;
    set<char> p={'2','3','7','8'};
    cout<<(!(p.count(s[s.size()-1])));
    return 0;
    }

  • 0
    @ 2015-08-04 11:16:29

    #include<cstdio>
    #include<cstring>
    int main()
    {
    char a[1000];
    int i,len,temp=0;
    gets(a);
    len=strlen(a);

    if( (a[len-1]-'0') %2==0 )
    printf("1\n");
    else printf("0\n");

    for(i=0;i<len;i++)
    {
    temp=temp+a[i]-'0';
    }
    if(temp%3==0)

    printf("1\n");
    else printf("0\n");

    if( ( (a[len-2]-'0') *10 + a[len-1]-'0')%4==0 )
    printf("1\n");
    else printf("0\n");

    if( ( (a[len-3]-'0') *100 + (a[len-2]-'0') *10 + a[len-1]-'0')%8==0 )
    printf("1\n");
    else printf("0\n");

    if( a[len-1]=='1' || a[len-1]=='4' || a[len-1]=='6' || a[len-1]=='9' || a[len-1]=='0' )
    printf("1");
    else printf("0");

    return 0;
    }

  • 0
    @ 2015-05-11 10:24:54

    为什么开35不行 破题 应该行啊

    #include<stdio.h>
    #include<string.h>
    int main()
    {
    char s[100003];
    int l,i,n[100030],f,sum,a4,a8;
    while(~scanf("%s",s))
    {
    f=0,sum=0,a4=a8=0;
    l=strlen(s);
    for(i=0;i<l;i++)
    {
    n[i]=s[i]-'0';
    a4=a4*10+n[i];
    a4%=4;
    a8=a8*10+n[i];
    a8%=8;
    sum+=n[i];
    }
    if(n[l-1]%2==0)
    {
    printf("1\n");
    f++;
    }
    else
    printf("0\n");
    if(sum%3==0)
    {
    printf("1\n");
    f++;
    }
    else
    printf("0\n");
    if(a4==0)
    {
    printf("1\n");
    f++;
    }
    else
    printf("0\n");
    if(a8==0)
    {
    printf("1\n");
    f++;
    }
    else
    printf("0\n");
    if(n[l-1]==0||n[l-1]==1||n[l-1]==4||n[l-1]==5||n[l-1]==6||n[l-1]==9)
    printf("1\n");
    else
    printf("0\n");
    }
    return 0;
    }

  • 0
    @ 2015-02-10 13:02:17

    首先,看了别人的题解,我知道了被4整除只需要看后2位,因为100肯定被4整除;
    被8整除只需要看后3位,因为1000肯定被8整除。
    而我做时却是用除法求余数。
    其次,完全平方数个位必然是0,1,4,5,6,9.而这道题不需要考虑这个,题的难度对于不知道此规律的人就减小了。
    而对于知道此规律的人难度反而加大了。
    #include<iostream>
    #include<string.h>
    #include<stdio.h>
    using namespace std;

    int a[10003];
    int size;
    int div(int k){
    int i;
    int d = 0;
    for (i = size; i >= 0; i--){
    d *= 10;
    d += a[i];
    d %= k;
    }
    return d;
    }
    int main(){
    freopen("in.txt", "r", stdin);

    char s[10003];
    cin >> s;
    int i;
    for (i = 0; s[i]; i++);
    size = i;
    for (i--; i >= 0; i--)a[size - i - 1] = s[i] - '0';
    if (a[0] % 2==0)cout << 1 << endl;
    else cout << 0 << endl;
    int sum = 0;
    for (i = 0; i < size; i++)sum += a[i];
    if (sum % 3 == 0)cout << 1 << endl;
    else cout << 0 << endl;
    if (div(4) != 0)cout << 0 << endl;
    else cout << 1 << endl;
    if (div(8) != 0)cout << 0 << endl;
    else cout << 1 << endl;
    cout << 1 << endl;
    return 0;
    }

  • 0
    @ 2014-10-22 19:30:19

    这题简直逗,f[5]恒为1……
    var a:array[1..10000] of longint;
    f:array[1..5] of longint;
    n,k,i,j,p,q:longint;
    s:ansistring;
    begin
    read(s);
    n:=length(s);
    for i:=1 to n do
    a[i]:=ord(s[n-i+1])-ord('0');
    if a[1] mod 2=0 then f[1]:=1 else f[1]:=0;
    k:=0;
    for i:=1 to n do
    begin
    k:=k+a[i];
    end;
    if k mod 3=0 then f[2]:=1 else f[2]:=0;
    p:=a[1]+10*a[2];
    if p mod 4=0 then f[3]:=1 else f[3]:=0;
    q:=a[1]+10*a[2]+100*a[3];
    if q mod 8 =0 then f[4]:=1 else f[4]:=0;

    f[5]:=1;
    for i:=1 to 5 do
    writeln(f[i]);
    end.

  • 0
    @ 2014-08-11 00:30:05

    坑题
    s[1] in [0,1,4,5,6,9]来判断完全平方
    program p1490;
    var a,b:array[0..31000] of longint;
    n,i,k,t,t1,t2:longint;
    s:ansistring;
    //
    function make(p:longint):longint;
    var i:longint;
    begin
    a:=b;
    for i:=a[0] downto 2 do
    begin
    a[i-1]:=a[i-1]+10*(a[i] mod p);
    a[i]:=a[i] div p;
    end;
    if a[1] mod p=0 then exit(1)
    else exit(0);
    end;
    //
    begin
    readln(s);
    a[0]:=length(s);
    for i:=1 to length(s) do a[i]:=ord(s[i])-ord('0');
    for i:=1 to a[0] div 2 do begin k:=a[i];a[i]:=a[a[0]+1-i];a[a[0]+1-i]:=k;end;
    b:=a;
    t:=make(2);t1:=t;writeln(t);
    t:=make(3);writeln(t);
    t:=make(4);t2:=t;writeln(t);
    t:=make(8);writeln(t);
    if b[1] in [0,1,4,5,6,9] then write(1)
    else write(0);
    end.

  • 0
    @ 2014-01-03 11:33:57

    #include <stdio.h>
    #include <string.h>

    char s[33];

    void dv(int x)
    {
    printf("%d\n" , x);
    }

    int main()
    {
    scanf("%s" , s);

    int ilen = strlen(s) , i , sum = 0;

    if(s[ilen - 1] % 2 == 0) dv(1);
    else dv(0);
    for(i = 0 ; i < ilen ; i ++ )
    {
    sum += s[i] - '0';
    }
    if(sum % 3 == 0) dv(1);
    else dv(0);
    if((s[ilen - 1] - '0' + (s[ilen - 2] - '0') * 10) % 4 == 0) dv(1);
    else dv(0);
    if((s[ilen - 1] - '0' + (s[ilen - 2] - '0') * 10 + (s[ilen - 3] - '0') * 100) % 8 == 0) dv(1);
    else dv(0);
    if(s[ilen - 1] == '0' || s[ilen - 1] == '1' || s[ilen - 1] == '4' || s[ilen - 1] == '9' || s[ilen - 1] == '6' || s[ilen - 1] == '5') dv(1);
    else dv(0);
    return 0;
    }
    如何秒杀?

  • 0
    @ 2014-01-03 11:33:24

    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 436 KiB, score = 20
    测试数据 #1: Accepted, time = 0 ms, mem = 436 KiB, score = 20
    测试数据 #2: Accepted, time = 15 ms, mem = 436 KiB, score = 20
    测试数据 #3: Accepted, time = 0 ms, mem = 440 KiB, score = 20
    测试数据 #4: Accepted, time = 0 ms, mem = 440 KiB, score = 20
    Accepted, time = 15 ms, mem = 440 KiB, score = 100

  • 0
    @ 2013-07-31 16:20:35

    AC的第90题

  • 0
    @ 2012-10-15 10:53:35

    BS出题人

    我开1..32的数组和string怎么改都过不去,最后一怒之下开了个1..10000的数组和ansistring结果AC了

    再写题的小心了!

    {

    ID:darkgod-z

    PROG:vijos P1490

    HANG:Pascal

    }

    var

    a:array [1..10000] of integer;

    i,sum,n:integer;

    s:ansistring;

    begin

    readln(s);

    n:=length(s);

    for i:=1 to n do

    a[i]:=ord(s[n-i+1])-ord('0');

    if a[1] mod 2=0 then writeln('1') else writeln('0');

    sum:=0;

    for i:=1 to n do inc(sum,a[i]);

    if sum mod 3=0 then writeln('1') else writeln('0');

    if (a[2]*10+a[1]) mod 4=0 then writeln('1') else writeln('0');

    if (a[3]*100+a[2]*10+a[1]) mod 8=0 then writeln('1') else writeln('0');

    writeln('1');

    end.

  • 0
    @ 2010-07-09 08:50:32

    强烈质疑这道题的正确性……

  • 0
    @ 2009-11-09 21:20:17

    纪念我的AC100题...

    那个可能的确Nice...

    可能,无限可能...

  • 0
    @ 2009-11-09 21:19:38

    晒晒~~

    var i,sum:longint;

    s:string;

    a:array[1..10000] of longint;

    procedure reed;

    var i,l:longint;

    begin

    readln(s);

    l:=length(s);

    sum:=0;

    for i:=1 to l do

    begin

    a[i]:=ord(s[i])-ord('0');

    sum:=sum+a[i];

    end;

    if a[l] mod 2 =0 then writeln(1) else writeln(0);

    if sum mod 3 =0 then writeln(1) else writeln(0);

    if (a[l-1]*10+a[l]) mod 4 =0 then writeln(1) else writeln(0);

    if (a[l-2]*100+a[l-1]*10+a[l]) mod 8=0 then writeln(1) else writeln(0);

    writeln(1);

    end;

    begin

    reed;

    end.

  • 0
    @ 2009-11-02 17:10:23

    楼主乃晴天一霹雳!!!!!!!!

    ---|---|---|---|---|---|---|---|---|---|---|---|-

    var i,a1,a2,a3,h:longint;s,t:string;

    begin

    readln(s);

    a1:=0;

    for i:=1 to length(s) do

    begin

    a1:=ord(s[i])-ord('0')+a1;

    end;

    t:=copy(s,length(s),1);

    val(t,h);

    if h mod 2=0 then writeln(1) else writeln(0);

    if a1 mod 3=0 then writeln(1) else writeln(0);

    t:=copy(s,length(s)-1,2);

    val(t,h);

    if h mod 4=0 then writeln(1) else writeln(0);

    t:=copy(s,length(s)-2,3);

    val(t,h);

    if h mod 8 =0 then writeln(1) else writeln(0);

    t:=copy(s,length(s),1);

    if (t='2')or(t='3')or(t='7')or(t='8') then writeln(0) else writeln(1);

    end.

    看完题就知道有鬼了

信息

ID
1490
难度
3
分类
数论 点击显示
标签
(无)
递交数
1091
已通过
568
通过率
52%
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