50—求助

P1009 [NOIP1998 普及组] 阶乘之和

saam @ 2024-08-06 20:17:14

#import "bits/stdc++.h"
#define ll long long
using namespace std;
ll sum = 1;
ll jc(ll n){
    ll sum=1;
    for(int i=2;i<=n;i++)sum*=i;
    return sum;
}
signed main(){
    ll n,cos=3;
    cin >> n;
    for(int i=3;i<=n;i++)cos+=jc(i);
    cout << cos;
    return 0;
}

by saam @ 2024-08-06 20:21:36

本人不会高精度计算?

by ptxy2352010111 @ 2024-08-06 21:50:23

不用高精度过不了，50^50超long long 了

by saam @ 2024-08-06 22:04:47

@ptxy2352010111 高精度是什么?

by GLr137 @ 2024-08-10 17:49:11

……高精度就是高精度，为什么呢？因为高精度是高精度，所以呢高精度是高精度（也就是高精度）。

by GLr137 @ 2024-08-10 17:50:07

之所以高精度是高精度，是因为高精度是高精度，所以代表高精度的高精度本质上还是高精度。

by GLr137 @ 2024-08-10 17:52:15

# include <iostream>
# include <cstring>
using namespace std;
int sz1[1000000] = {0}, sz2[1000000] = {0}, sz3[1500000] = {0}, len, x;
string hs(string s1, string s2){
    string s3 = "";
    memset(sz1, 0, sizeof(sz1));
    memset(sz2, 0, sizeof(sz2));
    memset(sz3, 0, sizeof(sz3));
    x = 0;
    for (int i = s1.size() - 1; i >= 0; i--){
        sz1[s1.size() - i - 1] = s1[i] - '0';
    }
    for (int i = s2.size() - 1; i >= 0; i--){
        sz2[s2.size() - i - 1] = s2[i] - '0';
    }
    if (s1.size() > s2.size()){
        len = s1.size();
    }else{
        len = s2.size();
    }
    for (int i = 0; i < len; i++){
        sz3[i] = (sz1[i] + sz2[i] + x) % 10;
        x = (sz1[i] + sz2[i] + x) / 10;
    }
    if (x == 1){
        s3 += x + 48;
    }
    for (int i = len - 1; i >= 0; i--){
        s3 += sz3[i] + 48;
    }
    return s3;
}
string hs2 (string s1, int s2){
    int y = 0;
    string s3, s4;
    s4 = s1;
    for (int i = 1; i <= s2 - 1; i++){
        s3 = hs(s1, s4);
        s1 = s3;
    }
    return s1;
}
int main (){
    int n, o = 1;
    string u = "1", v = "0", w = "0", sz[55] = {"1"};
    cin >> n;
    for (int i = 1; i <= n; i++){
        u = hs2(sz[i - 1], i);
        sz[i] = u;
    }
    for (int i = 1; i <= n; i++){
        w = hs(w, sz[i]);
    }
    cout << w;
    return 0;
}

这是高精度“乘”法（本质上是高精度加法，是一层一层累加起来的，也就是高精度乘上单精度。

by GLr137 @ 2024-08-10 17:55:35

求关

by GLr137 @ 2024-08-10 17:56:03

)

by Kete @ 2024-08-19 13:42:16

废话文学 @saam

我的文章里写了高精度

```cpp #include<bits/stdc++.h> using namespace std; int main() { int n,i,j,a[1111]= {0},b[1111]= {0}; scanf("%d", &n);//输入 a[0]=b[0]=1;//首位初始化 for (i=2; i<=n; i++) {//一的阶乘到n的阶乘 for (j=0; j<100; j++) { //遍历 b[j]*=i;//阶乘 } for (j=0; j<100; j++)//遍历 if (b[j]>9) {//进位处理 b[j+1] += b[j]/10;//下一位是进十位 b[j]%=10;//留个位 } for (j=0; j<100; j++) {//遍历 a[j]+=b[j];//加上上一次阶乘的结构 if (a[j]>9) {//进位处理 a[j+1]+=a[j]/10;//下一位是进十位 a[j]%=10;//留个位 } } } for (i=100; i>=0&&a[i]==0; i--);//通过循环来剥离出合适的答案 for (j=i; j>=0; j--) printf("%d", a[j]);//输出答案 return 0; } ``` ------------ $$The+End$$

by saam @ 2024-08-22 22:27:56

谢谢

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