23 条题解

  • 14
    @ 2016-07-27 08:54:23

    n>m时
    n!=1*2*.......*m*...*n;
    n! % m =0
    所以

    if(n>=m)n=m;

    • @ 2017-03-03 22:56:22

      orz,厉害的算法!!!瞬间AC!!!

  • 5
    @ 2017-06-03 15:59:06
    #include<iostream>
    using namespace std;
    
    int main(){
        long long t=1,n,m,ans=0;
        cin >>n>>m;
        if (n>m) n=m;
        for (long long i=1;i<=n;++i)
            t=(t*i)%m,ans=(ans+t)%m;
        cout <<ans<<endl;
        return 0;
    }
    
  • 0
    @ 2018-10-04 16:58:38

    p党来了
    var
    a,b,s,j:int64;
    i:longint;
    begin
    read(a,b);
    if a>b then a:=b-1;
    s:=1;
    for i:=1 to a do
    begin
    s:=s*i;
    s:=s mod b;
    j:=j+s;
    end;
    write(j mod b);
    end.

  • 0
    @ 2018-08-20 09:24:44
    /*
    做这题是在做比赛题目的时候,
    看到这题数据范围,看到n这么大,直接哭了出来
    然后就去看后面的题目了
    果然我太弱了瞬间发现后面的也不会
    然后一路暴力回来
    突然脑子一动233333
    发现如果计算的n,m
    n比m会更大,那么计算到m!,(m+1)!,(m+2)!的时候
    一取模就没了啊
    因为里面已经含有了m这个因子
    所以余数一定是0
    就可以不用考虑
    然后就用的是对1到min(n,m)计算阶乘然后相加
    发现诶不对这个复杂度明显超时啊
    100万的范围估计就是O(n)?
    突然发现自己傻逼了
    我们计算1到min(n,m)的阶乘的时候
    比如计算i的阶乘,那么i+1的阶乘就可以不用直接算了
    就直接在保留的i的阶乘上乘上一个i+1就好了
    以此类推
    好像是一个很简单的技巧只能说做题目太少了
    不过还好是想出来了
    */
    #include <iostream>
    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    #include <iomanip>
    #include <cstdlib>
    using namespace std;
    
    long long n,m;
    int sum=1,ans;
    
    int main()
    {
        cin>>n>>m;
        for (long long i=1;i<=min(n,m);i++)
        {
            sum=(i*sum)%m;
            ans=(ans+sum)%m;
        }
        cout<<ans%m<<endl;
        return 0;
    }
         
    
  • 0
    @ 2016-11-15 11:36:38

    数据有毒
    编译成功

    测试数据 #0: WrongAnswer, time = 0 ms, mem = 560 KiB, score = 0
    测试数据 #1: Accepted, time = 0 ms, mem = 556 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 556 KiB, score = 10
    测试数据 #3: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 552 KiB, score = 10
    测试数据 #5: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    测试数据 #6: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    测试数据 #7: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    测试数据 #8: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    测试数据 #9: Accepted, time = 15 ms, mem = 556 KiB, score = 10
    WrongAnswer, time = 90 ms, mem = 560 KiB, score = 90

    九次WA了。。。

    • @ 2017-07-03 16:35:40

      我也是第一个死也过不了

  • 0
    @ 2015-10-10 14:51:04

    ###pascal code
    program P1964;
    var ans,i:longint;
    sum,n,m:int64;
    begin
    read(n,m); sum:=1; if n>1000000 then n:=m;
    for i:=1 to n do
    begin
    sum:=sum*i mod m; ans:=(ans+sum) mod m;
    end;
    write(ans);
    end.

  • 0
    @ 2015-10-04 17:53:11
    #include <iostream>
    using namespace std;
    
    int main()
    {
    
        long long n,m;
        long long sum=0,temp=1;
        cin>>n>>m;
        for(int i=1;i<=min(n,m);i++)
        {
            temp*=i;
            temp%=m;
            sum+=temp;
            sum%=m;
        }
        cout<<sum<<endl;
        
        return 0;
    }
    

    终于ac。。。

  • 0
    @ 2015-10-04 10:27:46

    测试数据 #8?

  • 0
    @ 2015-10-03 23:15:27

    Block code

    #include<cstdio>

    int main()
    {
    long long n,m,t=1,ans=0;
    scanf("%lld%lld",&n,&m);
    if(n>=m) n=m-1;
    for(int i=1;i<=n;i++)
    {
    t=(t*i)%m;
    ans+=t;

    }
    ans%=m;
    printf("%lld",ans);
    return 0;

    }
    用一个变量累成×,再一个变量累加就好了,累×变量实时取模,累加变量用long long,最后一次取模节省时间
    当n>=m时,n! % m==0,所以后面的全部忽略掉,累加到(m-1)!过

  • 0
    @ 2015-10-03 22:34:40

    #include <cstdio>
    int main()
    {
    long long n;
    int m;
    scanf("%I64d %d", &n, &m);
    if (n > m)
    n = m;
    int ans = 0;
    int temp = 1;
    for (int i = 1; i <= n; i++)
    {
    temp = temp * i % m;
    ans = (ans + temp) % m;
    }
    printf("%d\n", ans);
    return 0;
    }

    为什么
    测试数据 #0: Accepted, time = 0 ms, mem = 444 KiB, score = 10
    测试数据 #1: Accepted, time = 0 ms, mem = 448 KiB, score = 10
    测试数据 #2: Accepted, time = 0 ms, mem = 444 KiB, score = 10
    测试数据 #3: Accepted, time = 15 ms, mem = 440 KiB, score = 10
    测试数据 #4: Accepted, time = 0 ms, mem = 448 KiB, score = 10
    测试数据 #5: Accepted, time = 0 ms, mem = 448 KiB, score = 10
    测试数据 #6: Accepted, time = 0 ms, mem = 440 KiB, score = 10
    测试数据 #7: Accepted, time = 0 ms, mem = 440 KiB, score = 10
    测试数据 #8: WrongAnswer, time = 15 ms, mem = 440 KiB, score = 0
    测试数据 #9: Accepted, time = 15 ms, mem = 448 KiB, score = 10
    WrongAnswer, time = 45 ms, mem = 448 KiB, score = 90

    • @ 2015-10-03 22:37:13

      temp * i有可能会爆int

  • 0
    @ 2015-10-03 22:26:37

    program alisky;
    var n,m,sum,tot,past,past1:qword;
    i:longint;
    begin
    readln(n,m);
    sum:=1;tot:=1;
    for i:=2 to n do
    begin
    if (i>=m) then break;
    tot:=tot*i mod m;
    sum:=(sum+tot) mod m;
    end;
    writeln(sum);
    end.
    @少女夜夜
    为什么这个第六组过不了

  • 0
    @ 2015-10-03 22:08:45

    大神看看哪错了?爆了第九个点
    #include<stdio.h>
    #include<iostream>
    using namespace std;
    int x[1010000];
    int main(){
    int m;
    long long n;
    scanf("%lld%d",&n,&m);
    if (n>m) n=m;
    x[0]=1;
    int sum=0;
    for (int i=1;i<=n;i++)
    {
    x[i]=(x[i-1]*i)%m;
    sum=(sum +x[i]) %m;
    }
    printf("%d\n",sum);
    }

  • -1
    @ 2016-12-08 20:04:19
    #include<cstdio>
    #include<cstring>
    using namespace std;  
    main(){
        int n,m;
        cin>>n>>m;
        int x=1;
        int d=0;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=i;j++){
                x=x*j;
            }
            d=x+d;
            x=1;
        }
        cout<<d;
        return 0;  
    }  
    
  • -1
    @ 2016-08-20 08:57:18

    {
    N<=1000000000000000000,M<=1000000

    比M大的直接忽略。 (因为 mod M)

    }
    ```PASCAL
    var
    i:longint;
    N,M,tmp,ans:int64;

    begin
    readln(N,M);
    if N>M then N:=M-1;
    tmp:=1;
    for i:=1 to N do
    begin
    tmp:=(tmp*i)mod M;
    if tmp=0 then break;
    ans:=(ans+tmp) mod M;
    end;
    writeln(ans);
    end.```

  • -1
    @ 2016-07-20 15:49:23

    一组大数据WA掉了,求大神帮忙看看!!!
    #include<iostream>
    using namespace std;
    long long m,n;
    int main ()
    {
    cin>>n>>m;
    int ans=0,sum=1;
    for (int i=1;i<=min(n,m);i++)
    {
    sum=(i*sum)%m;
    ans=(ans+sum)%m;
    }
    cout<<ans;
    return 0;
    }

  • -1
    @ 2016-07-20 15:34:04

    求指教,后面四组大数据过不了,怎么解决?
    #include <iostream>
    using namespace std;
    int m,n;

    int main ()
    {
    cin>>n>>m;
    int ans=0,sum=1;
    for (int i=1;i<=n;i++)
    {
    sum=(i*sum)%m;
    ans=(ans+sum)%m;
    }
    cout<<ans;
    return 0;
    }

  • -1
    @ 2016-03-29 12:48:29

    希望大家帮忙解答一下,这段代码后四个全都WA

    import java.util.Scanner;
    java
    class Main
    {
    public static void main(String[] args)
    {
    Scanner in= new Scanner(System.in);
    int n = in.nextInt();
    int m = in.nextInt();
    long difficulty = 1;
    long total=0;
    int i = 0;
    while (i!=n)
    {
    i++;
    difficulty = difficulty*i%m;
    total =(total+ difficulty)%m;
    }
    System.out.println(total);
    }
    }

  • -1
    @ 2015-11-05 19:55:59

    根据mod的性质,只要在阶乘和中找到一个和 mod m=0,那么之后的数据都可以舍去,因为 m! mod m=0 所以最大就找到m。
    #include <iostream>
    using namespace std;
    long long int n,m,s=0,t=1;
    int main()
    {
    cin>>n>>m;
    if(n>=m)n=m;
    for(long long int i=1;i<=n;i++)
    {
    t=(t*i)%m;
    s=(s+t)%m;
    }
    cout<<s%m;
    return 0;
    }

  • -1
    @ 2015-10-25 09:47:25

    /*

    Author : Slience_K
    Belong : C++
    Pro : Vijos P 1964

    */
    #include <cstdio>
    #include <algorithm>
    #define LL long long
    using namespace std;
    LL n , m , ans , sum;
    int main(){
    freopen( "P1964.in" , "r" , stdin );
    scanf( "%I64d%I64d" , &n , &m );
    n = min( n , m - 1 );
    sum = 1;
    ans = 0;
    for( int i = 1 ; i <= n ; i++ ){
    sum *= i;
    sum %= m;
    ans = ( ans + sum ) % m;
    }
    printf( "%I64d" , ans );
    return 0;
    }

  • -1
    @ 2015-10-07 18:24:58

    #include<iostream>
    int main()
    {
    unsigned long long n;
    long m;
    std::cin>>n>>m;
    long long ans(1);
    long long cnt(1);
    for(long i=2;cnt!=0&&i<=n;i++)
    {
    cnt=(cnt*i)%m;
    ans=(ans+cnt)%m;
    }
    std::cout<<ans;
    return 0;
    }

信息

ID
1964
难度
6
分类
d 点击显示
标签
(无)
递交数
2223
已通过
519
通过率
23%
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