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题解

54 条题解

  • 2
    @ 2018-02-12 21:43:15
    #include <cmath>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    #include <iomanip>
    #include <iostream>
    #include <algorithm>
    #include <vector>
    #include <deque>
    #include <set>
    #include <limits>
    #include <string>
    #include <sstream>
    #include <thread>
    using namespace std;
    
    const int oo_min=0xc0c0c0c0,oo_max=0x3f3f3f3f;
    
    namespace dts
    {
        typedef long long ll;
        
        ll n,a,b;
        ll rec[15+1];
        
        ll calc(ll a,ll b,ll key)
        {
            ll l,r;
            if (a%key!=0)
                l=((a/key)+1)*key;
            else
                l=a;
            if (b%key!=0)
                r=(b/key)*key;
            else
                r=b;
            if (l<=r)
                return (r-l)/key+1;
            else
                return 0;
        }
        
        ll gcd(ll a,ll b)
        {
            if (b==0)
                return a;
            else
                return gcd(b,a%b);
        }
        
        ll lcm(ll a,ll b)
        {
            return a*b/gcd(a,b);
        }
        
        void dfs(ll pos,ll num,ll sig,ll *ans)
        {
            *ans+=sig*calc(a,b,num);
            for (ll i=pos+1;i<=n;i++)
                dfs(i,lcm(num,rec[i]),-sig,ans);
        }
        
        ll work()
        {
            ll temp=8,ans=0;
            dfs(0,temp,1,&ans);
            return ans;
        }
        
        void main()
        {
            while (~scanf("%lld",&n))
            {
                for (ll i=1;i<=n;i++)
                    scanf("%lld",&rec[i]);
                scanf("%lld%lld",&a,&b);
                printf("%lld\n",work());
            }
        }
    };
    
    int main()
    {
        dts::main();
    }
    
  • 1
    @ 2017-02-09 18:57:38

    水题
    C++
    #include<cstdio>
    #include<iostream>
    #define LL long long
    using namespace std;
    LL n,num[20],a,b,ans;
    LL lcm(LL a,LL b,LL x,LL y)
    {
    if(b==0)
    return x*y/a;
    return lcm(b,a%b,x,y);
    }
    void dfs(LL i,LL sum,LL cnt)
    {
    if(i>n)
    {
    if(cnt%2)
    ans-=(b/sum-a/sum);
    else
    ans+=(b/sum-a/sum);
    return;
    }
    dfs(i+1,sum,cnt);
    dfs(i+1,lcm(sum,num[i],sum,num[i]),cnt+1);
    }
    int main()
    {
    scanf("%I64d",&n);
    for(int i=1;i<=n;i++)
    scanf("%I64d",&num[i]);
    scanf("%I64d%I64d",&a,&b);
    dfs(1,8,0);
    printf("%I64d",ans);
    return 0;
    }

  • 0
    @ 2017-03-11 15:31:11

    尽量易懂

    #include<cstdio>
    #include<algorithm>
    #define LL long long
    using namespace std;
    const int N=25;
    LL n,a,b;
    LL ai[N];
    int Gcd(int a,int b){
    return b==0?a:Gcd(b,a%b);
    }
    void Dfs(LL k,LL las,LL &sum,LL num){
    int pos=num;
    for(int i=las+1;i<=n;i++){
    int g=Gcd(num,ai[i]);
    if (num/g*ai[i]>b) continue;

    else{
    num=num/g*ai[i];
    if (k&1) sum+=(b/num)-(a/num);
    else sum-=(b/num)-(a/num);
    Dfs(k+1,i,sum,num);
    num=pos;
    }
    }
    return;

    }
    int main(){
    scanf("%lld",&n);
    for(int i=1;i<=n;i++){
    scanf("%lld",&ai[i]);
    if (ai[i]==1||ai[i]==2||ai[i]==4||ai[i]==8){
    printf("0");
    return 0;

    }

    }
    scanf("%lld %lld",&a,&b);
    LL sum=0;
    Dfs(1,0,sum,8);
    printf("%lld",(b/8)-(a/8)-sum);
    return 0;

    }

  • 0
    @ 2016-11-03 21:15:47

    //题解真是难懂(看来要去学语文了。),不如自己去想。

    奇数的减去 偶数的加上 。 不要忘了8 //前3个点似乎不用8...
    Dfs 枚举那几个数 Gcd 求最大公约数来求最小公倍数。

    var
        i,n,a,b,dis,point:longint;
        ans:int64;
        data,g:array[0..15]of int64;
        flag:array[0..10000]of boolean;
    
    function gcd(a,b:int64):int64;
    begin
        if b=0 then exit(a) else
            exit(gcd(b,a mod b));
    end;
    
    procedure check(dis:longint);
    var
        i:longint;
        tmp1,tmp2:int64;
    begin
        tmp1:=g[1]*8;
        tmp2:=tmp1 div gcd(g[1],8);
        for i:=2 to dis do
        begin
            tmp1:=tmp2*g[i];
            tmp2:=tmp1 div gcd(tmp2,g[i]);
            if tmp2>b then exit;
        end;
        if (dis and 1=1) then ans:=ans-((b div tmp2)-((a-1)div tmp2)) else
            ans:=ans+((b div tmp2)-((a-1)div tmp2));
    
    end;
    
    procedure Dfs(head,dep,dis:longint);
    var
        i:longint;
    begin
        if dep=dis+1 then
        begin
            check(dis);
            exit;
        end;
        if head>n then exit;
        for i:=head to n do
        begin
            g[dep]:=data[i];
            Dfs(i+1,dep+1,dis)
        end;
    end;
    
    begin
        readln(n);
        for i:=1 to n do
        begin
            inc(point);
            read(data[point]);
            if not flag[data[point]] then flag[data[point]]:=true else
                dec(point);
        end;
        n:=point;
        readln(a,b);
        ans:=(b div 8)-((a-1) div 8);
        for dis:=1 to n do
            Dfs(1,1,dis);
        writeln(ans);
    end.
    
  • 0
    @ 2016-08-29 16:51:05

    一不做二不休,把代码改成纯函数式的,从宏观层面毫无副作用。。
    我是函数式编程的脑残粉。。
    ```c++
    #include <bits/stdc++.h>
    using namespace std;

    typedef long long LL;

    inline LL gcd(LL a, LL b)
    {
    return b == 0 ? a : gcd(b, a%b);
    }

    inline LL lcm(LL a, LL b)
    {
    return a/gcd(a, b)*b;
    }

    inline LL num(LL a, LL b, LL c)
    {
    return a%c?b/c-a/c:b/c-a/c+1;
    }

    inline LL two_num(LL a, LL b, LL c, LL d)
    {
    return max(0ll, num(a, b, c) - num(a, b, lcm(c,d)));
    }

    LL dfs(LL li[], LL i, LL n, LL m, LL get, LL a, LL b)
    {
    if (i == n && m == 0)
    return two_num(a, b, 8, get);
    if (i == n || m < 0)
    return 0;
    return dfs(li, i+1, n, m, get, a, b) + dfs(li, i+1, n, m-1, lcm(get, li[i]), a, b);
    }

    LL list_num(int i, int k, LL a, LL b, LL c[], LL n)
    {
    if (i == n+1)
    return 0;
    return k*dfs(c, 0, n, i, 1, a, b) + list_num(i+1, -k, a, b, c, n);
    }

    void read(LL a[], LL i)
    {
    if (i == -1)
    return;
    read(a, i-1);
    cin >> a[i];
    }

    int main()
    {
    LL n, a[20], x, y;
    cin >> n;
    read(a, n-1);
    cin >> x >> y;
    cout << list_num(1, 1, x, y, a, n) << endl;
    return 0;
    }
    ```

  • 0
    @ 2016-05-11 23:59:02

    #include <stdio.h>
    #include <algorithm>
    #define ll long long
    using namespace std;
    int sum=0,n,l;
    ll f[16],st,ed;
    ll gcd (ll a,ll b)
    {if (a%b==0) {return b;}
    return gcd(b,a%b);
    }
    ll z (ll a,ll b)
    {if (a<b) {ll t=a;a=b;b=t;}
    return a*b/gcd(a,b);
    }
    void zh (int re,int pl,ll ng)
    {if (re==0)
    {int k=(ed-st+1)/ng;
    if ((ed%ng)<((st-1)%ng)) {k++;}
    sum+=k*l;
    return;
    }
    if (ng>ed) {return;}
    if (pl>n) {return;}
    int i;
    for (i=pl;i<=n;i++)
    {zh(re-1,i+1,z(ng,f[i]));}
    return;
    }
    int main (){
    int i;
    scanf ("%d",&n);
    for (i=1;i<=n;i++)
    {scanf ("%I64d",&f[i]);}
    sort (f+1,f+n+1);
    scanf ("%I64d%I64d",&st,&ed);
    sum=(ed-st+1)/8;
    if (ed%8<(st-1)%8) {sum++;}
    l=-1;
    for (i=1;i<=n;i++)
    {zh(i,1,8);l*=(-1);}
    printf ("%d\n",sum);
    return 0;
    }

  • 0
    @ 2015-10-14 16:16:37

    个人觉得代码写的还算清晰明了=_=

    ##
    #include<iostream>
    using namespace std;

    typedef long long LL;
    LL n,L,R,A[20];

    inline LL gcd(LL a,LL b) {
    if(!b) return a;
    return gcd(b,a%b);
    }
    inline LL lcm(LL a,LL b) {
    return a*b/gcd(a,b);
    }

    int main() {
    cin>>n;
    for(int i=0;i<n;i++) cin>>A[i];
    cin>>L>>R;
    LL ans=R/8-(L-1)/8;
    for(int s=1;s<=(1<<n)-1;s++)
    {
    LL _lcm=8,cnt=0;
    for(int i=0;i<n;i++) if(s&(1<<i)) _lcm=lcm(_lcm,A[i]) , cnt++;
    if(cnt&1) ans -= R/_lcm-(L-1)/_lcm;
    else ans += R/_lcm-(L-1)/_lcm;;
    }
    cout<<ans;
    return 0;
    }

  • 0
    @ 2015-07-31 15:59:33

    记录信息
    评测状态 Accepted
    题目 P1629 八
    递交时间 2015-07-31 15:55:29
    代码语言 C++
    评测机 VijosEx
    消耗时间 2138 ms
    消耗内存 509332 KiB
    评测时间 2015-07-31 15:55:34
    评测结果
    编译成功

    测试数据 #0: Accepted, time = 0 ms, mem = 509332 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 509332 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 509332 KiB, score = 10
    测试数据 #3: Accepted, time = 234 ms, mem = 509332 KiB, score = 10
    测试数据 #4: Accepted, time = 125 ms, mem = 509332 KiB, score = 10
    测试数据 #5: Accepted, time = 281 ms, mem = 509332 KiB, score = 10
    测试数据 #6: Accepted, time = 359 ms, mem = 509332 KiB, score = 10
    测试数据 #7: Accepted, time = 343 ms, mem = 509332 KiB, score = 10
    测试数据 #8: Accepted, time = 531 ms, mem = 509332 KiB, score = 10
    测试数据 #9: Accepted, time = 265 ms, mem = 509328 KiB, score = 10
    Accepted, time = 2138 ms, mem = 509332 KiB, score = 100
    代码
    #include <iostream>
    #include <stdio.h>
    #include <algorithm>
    using namespace std;
    int dive[200];
    int num[130000050];
    int main()
    {
    int n,ans=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
    scanf("%d",&dive[i]);
    for(int j=1;j<=3;j++)
    {
    if(dive[i]%2==0)dive[i]/=2;
    else break;
    }
    }
    sort(dive+1,dive+n+1);
    int a,b;
    scanf("%d%d",&a,&b);
    a=(a+7)/8;b/=8;
    for(int i=1;i<=n;i++)
    {
    int now=a/dive[i];
    if(now%dive[i]!=0)now++;
    for(int j=now;j*dive[i]<=b;j++)
    num[j*dive[i]]=1;
    }
    for(int i=a;i<=b;i++)
    if(!num[i])ans++;
    printf("%d",ans);
    }

    • @ 2015-07-31 16:02:16

      ###用的是 int ,占用内存较大 ,但我压缩了一下 ,少循环八分之七次 ,还是AC了

  • 0
    @ 2014-11-03 20:11:28

    P1629八
    Accepted

    记录信息

    评测状态 Accepted
    题目 P1629 八
    递交时间 2014-11-03 20:10:16
    代码语言 C++
    评测机 上海红茶馆
    消耗时间 995 ms
    消耗内存 122872 KiB
    评测时间 2014-11-03 20:10:18

    评测结果

    编译成功

    测试数据 #0: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

    测试数据 #1: Accepted, time = 78 ms, mem = 122872 KiB, score = 10

    测试数据 #2: Accepted, time = 78 ms, mem = 122868 KiB, score = 10

    测试数据 #3: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

    测试数据 #4: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

    测试数据 #5: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

    测试数据 #6: Accepted, time = 125 ms, mem = 122868 KiB, score = 10

    测试数据 #7: Accepted, time = 109 ms, mem = 122872 KiB, score = 10

    测试数据 #8: Accepted, time = 140 ms, mem = 122864 KiB, score = 10

    测试数据 #9: Accepted, time = 93 ms, mem = 122864 KiB, score = 10

    Accepted, time = 995 ms, mem = 122872 KiB, score = 100

    代码

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>

    using namespace std;

    bool x[125000000 + 10];
    int y[15 + 2];
    int i , j;
    int a , b;
    int n;
    long long k , l;
    long long sum;

    int main()
    {
    while( scanf( "%d" , &n ) != EOF )
    {
    sum = 0;
    memset( x , 1 , sizeof( x ) );
    memset( y , 0 , sizeof( y ) );
    for( i = 0 ; i < n ; i++ )
    scanf( "%d" , &y[i] );
    scanf( "%d %d" , &a , &b );
    if( a % 8 == 0 )
    a /= 8;
    else
    a = a / 8 + 1;
    b /= 8;
    for( i = 0 ; i < n ; i++ )
    {
    if( y[i] % 8 == 0 )
    y[i] /= 8;
    else
    while( y[i] % 2 == 0 )
    y[i] /= 2;
    k = y[i];
    if( a % k == 0 )
    l = a;
    else
    l = ( a / k + 1 ) * k;
    for( j = l ; j <= b ; j += k )
    if( x[j] )
    {
    x[j] = 0;
    sum++;
    }
    }
    cout << b - a + 1 - sum << endl;
    }
    return 0;
    }

    非容斥AC

  • 0
    @ 2014-09-08 15:42:56

    #include <iostream>
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    #include <math.h>

    using namespace std;

    bool x[1000000000 + 10];
    int y[15 + 2];
    int i , j;
    int a , b;
    int n;
    long long sum;

    int main()
    {
    while( cin >> n )
    {
    sum = 0;
    memset( y , 0 , sizeof( y ) );
    for( i = 0 ; i < n ; i++ )
    scanf( "%d" , &y[i] );

    scanf( "%d %d" , &a , &b );
    for( i = 0 ; i < n ; i++ )
    {
    if( a % y[i] == 0 )
    j = a / y[i];
    else
    j = ( ( a / y[i] ) + 1 ) * y[i];
    for( ; j <= b ; j += y[i] )
    x[j] = 1;
    }
    if( a % 8 == 0 )
    i = a / 8;
    else
    i = ( ( a / 8 ) + 1 ) * 8;
    for( ; i <= b ; i += 8 )
    if( x[i] == 0 )
    sum++;
    cout << sum << endl;
    memset( x , 0 , sizeof( x ) );
    }
    return 0;
    }

    • @ 2014-09-08 15:43:26

      为什么会MLE。。。应该只用不到120MB呀

  • 0
    @ 2014-08-09 23:41:56

    program p1629;
    var a:array[1..15] of int64;
    p1,p2,sum:int64;
    i,n:longint;
    //
    function make(p:int64):int64;
    begin
    make:=(p2 div p)-(p1 div p);
    end;
    //
    function gcd(p,l:int64):int64;
    begin
    if l=0 then exit(p)
    else gcd:=gcd(l,p mod l);
    end;
    //
    procedure dfs(i,k,l:longint;p:int64);
    var j:longint;
    f:int64;
    begin
    if i=k then
    begin
    if k mod 2=0 then sum:=sum+make(p)
    else sum:=sum-make(p);
    exit;
    end;
    if l+(i-k)>n then exit;
    for j:=l+1 to n do
    begin
    f:=a[j]*p div gcd(a[j],p);
    dfs(i,k+1,j,f);
    end;
    end;
    //
    begin
    readln(n);
    for i:=1 to n do read(a[i]);readln;
    read(p1,p2);
    sum:=make(8);
    for i:=1 to n do dfs(i,0,0,8);
    write(sum);
    end.

  • 0
    @ 2010-03-11 23:49:07

    #include

    #include

    #include

    __int64 n,left,right;

    __int64 num[20];

    __int64 a[20],b[20],p;//a数组,保存每个和8最小公倍数值 b数组保存选取情况,p保存有的数量 1取,0不取

    __int64 ans,fuck;

    __int64 xx[20];//保存计算的东东

    __int64 gcd(__int64 a,__int64 b){

    if (b == 0) return a;

    return gcd( b , a % b );

    }

    void print(){

    int i;

    for ( i = 1 ; i right) return;

    }

    c = right / temp - ( left - 1 ) / temp;

    if (fuck % 2 == 0)

    ans + = c;

    else ans - = c;

    }

    void make(){

    int i,c,temp;

    fuck = 0;//目前读取到0个数字

    for (i = 1 ; i < = p ; i ++) if (b[i] == 1) fuck++; //看看有几个数字

    if (fuck == 0) return ;

    // print();

    for (i = 1 ; i < = p ; i ++)

    if (b[i] == 1) get(i);//第i个数字,作为和8的最小公倍数

    }

    void dfs(int k){

    int i,j;

    // printf("%d ",k);

    if (k == p) { make(); return;} //满足条件时候,进行最后处理

    b[k+1] = 1;

    dfs(k + 1);

    b[k+1] = 0;

    dfs(k + 1);

    }

    void init(){

    int i;

    p=0;

    ans=0;

    memset(num,0,sizeof(num));

    memset(a,0,sizeof(a));

    memset(b,0,sizeof(b));

    scanf("%d",&n);//有n个数字

    for (i = 1 ; i < = n ; i ++) scanf("%d",&num[i]); //读入这n个数字

    scanf("%d%d",&left,&right); //读入左边和右边的边界条件

    for (i = 1 ; i < = n ; i ++){ //判断,并且构造每个数字和8的最小公倍数

    if (num[i] ==1 || num[i] ==2 || num[i] ==4 || num[i] ==8 ){

    // printf("0\n");

    return ;

    }

    a[++p]= ( num[i] * 8 ) / gcd( num[i] , 8 );

    }

    // printf("%d!",p);

    ans=right / 8 - ( left - 1 ) / 8;//初始化

    dfs(0);

    printf("%I64d\n",ans);

    }

    int main(){

    //---|main---|-

    init();

    system("pause");

    return 0;

    }

    只过3个点,哪个大牛帮忙看下……

  • 0
    @ 2009-11-09 17:55:29

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

    一开始想多了.....

  • 0
    @ 2009-11-09 12:34:41

    有没有不用 斥容 的

    得 高分的??

    • @ 2014-11-03 20:11:17

      P1629八
      Accepted

      记录信息

      评测状态 Accepted
      题目 P1629 八
      递交时间 2014-11-03 20:10:16
      代码语言 C++
      评测机 上海红茶馆
      消耗时间 995 ms
      消耗内存 122872 KiB
      评测时间 2014-11-03 20:10:18

      评测结果

      编译成功

      测试数据 #0: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

      测试数据 #1: Accepted, time = 78 ms, mem = 122872 KiB, score = 10

      测试数据 #2: Accepted, time = 78 ms, mem = 122868 KiB, score = 10

      测试数据 #3: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

      测试数据 #4: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

      测试数据 #5: Accepted, time = 93 ms, mem = 122872 KiB, score = 10

      测试数据 #6: Accepted, time = 125 ms, mem = 122868 KiB, score = 10

      测试数据 #7: Accepted, time = 109 ms, mem = 122872 KiB, score = 10

      测试数据 #8: Accepted, time = 140 ms, mem = 122864 KiB, score = 10

      测试数据 #9: Accepted, time = 93 ms, mem = 122864 KiB, score = 10

      Accepted, time = 995 ms, mem = 122872 KiB, score = 100

      代码

      #include <iostream>
      #include <stdio.h>
      #include <string.h>
      #include <algorithm>
      #include <math.h>

      using namespace std;

      bool x[125000000 + 10];
      int y[15 + 2];
      int i , j;
      int a , b;
      int n;
      long long k , l;
      long long sum;

      int main()
      {
      while( scanf( "%d" , &n ) != EOF )
      {
      sum = 0;
      memset( x , 1 , sizeof( x ) );
      memset( y , 0 , sizeof( y ) );
      for( i = 0 ; i < n ; i++ )
      scanf( "%d" , &y[i] );
      scanf( "%d %d" , &a , &b );
      if( a % 8 == 0 )
      a /= 8;
      else
      a = a / 8 + 1;
      b /= 8;
      for( i = 0 ; i < n ; i++ )
      {
      if( y[i] % 8 == 0 )
      y[i] /= 8;
      else
      while( y[i] % 2 == 0 )
      y[i] /= 2;
      k = y[i];
      if( a % k == 0 )
      l = a;
      else
      l = ( a / k + 1 ) * k;
      for( j = l ; j <= b ; j += k )
      if( x[j] )
      {
      x[j] = 0;
      sum++;
      }
      }
      cout << b - a + 1 - sum << endl;
      }
      return 0;
      }

      非容斥AC

  • 0
    @ 2009-11-09 11:41:30

    囧~

    通过   250人

  • 0
    @ 2009-10-24 23:09:09

    容斥...

    跳格被无情的吃了,晕……

    #include

    #include

    #include

    #define maxn 16

    long N,a,b,A[maxn],sign=-1,Ans;

    void swap(long long *a,long long *b){

    long long tmp=*a;*a=*b;*b=tmp;

    }

    long long gcd(long long a,long long b){

    if (!b) return a;

    if (a

  • 0
    @ 2009-10-19 19:02:06

    ...

  • 0
    @ 2009-10-07 10:20:40

    编译通过...

    ├ 测试数据 01:答案正确... 0ms

    ├ 测试数据 02:答案正确... 0ms

    ├ 测试数据 03:答案正确... 0ms

    ├ 测试数据 04:答案正确... 0ms

    ├ 测试数据 05:答案正确... 0ms

    ├ 测试数据 06:答案正确... 0ms

    ├ 测试数据 07:答案正确... 0ms

    ├ 测试数据 08:答案正确... 0ms

    ├ 测试数据 09:答案正确... 0ms

    ├ 测试数据 10:答案正确... 0ms

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

    Accepted 有效得分:100 有效耗时:0ms

    。。。。莫名其妙的30分。。。。

    我的上午就这么没了..=.=

  • 0
    @ 2009-10-05 22:24:57

    为什么!!!!!!!

    为什么设置long long的时候I64不行、 lld却可以啊!!!!!!!!

  • 0
    @ 2009-10-03 10:47:04

    Flag    Accepted

    题号   P1629

    类型(?)   数论 / 数值

    通过   200人

    提交   863次

    通过率   23%

    难度   1

    紧跟楼下的。。

    8月30日我做了一下这题。。只过了3个点。。

    现在终于不一样了

信息

ID
1629
难度
7
分类
组合数学 | 容斥原理数论 点击显示
标签
(无)
递交数
2374
已通过
476
通过率
20%
被复制
4
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