就对一半，不用高精度就没法对吗……

huboxuan @ 2024-07-15 23:39:23

#include <bits/stdc++.h>
using namespace std;
int n;
long long s;
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        long long factor=1;
        for(int j=1;j<=i;j++)
            factor*=j;
        s+=factor;
    }
    printf("%lld",s);
    return 0;
}

by Steve_xh @ 2024-07-16 00:15:38

@huboxuan 是

by OldDriverTree @ 2024-07-16 01:00:42

@huboxuan n 较大时显然会爆 long long，所以必须要写高精度

by luoxiaoran62 @ 2024-07-16 07:55:12

@huboxuan 这题要高精加和高精乘

by Max_hmt @ 2024-07-18 20:07:56

@huboxuan 这题数据较大，long long都爆，除了高精度我还没有想出别的办法

by Xuhappy @ 2024-07-22 16:32:20

@huboxuan 怎么解决的？求