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

if(n>=m)n=m;

#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;
}

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.

/*
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;
}
     

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。。。

###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.

#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。。。

测试数据 #8?

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

#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

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

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);
}

#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;  
}  

{
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.```

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;
}

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;
}

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);
}
}


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;
}

/*

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;
}

#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;
}