0
已解决
高级天翼
题目描述
用高精度计算出S=1!+2!+3!+…+n!(n≤50)S=1!+2!+3!+…+n! (n≤50)S=1!+2!+3!+…+n!(n≤50)
其中“!”表示阶乘，例如：5!=5×4×3×2×15!=5 \times 4 \times 3 \times 2 \times 15!=5×4×3×2×1。
输入格式
一个正整数NNN。
输出格式
一个正整数SSS，表示计算结果。
输入输出样例
输入 #1
3
输出 #1
9
说明/提示
注，《深入浅出基础篇》中使用本题作为例题，但是其数据范围只有 n≤20n\le20n≤20，使用书中的代码无法通过本题。
如果希望通过本题，请继续学习第八章高精度的知识。
#include<iostream>
using namespace std;
long long jc(int n)
{
long long a=1;
for(int i=1;i<=n;i++)
{
a*=i;
}
return a;
}
int main()
{
int n;
long long sum=0;
cin>>n;
for(int i=1;i<=n;i++)
{
sum+=jc(i);
}
cout<<sum;
return 0;
}
本来想氵一题的，但发现在酷町堂里AC了，但到洛谷就WA了，两个点
张恩泽在2020-12-23 18:40:04追加了内容
复制不了的话，QQ有一个截图识字的功能【疯狂明示】
张恩泽在2020-12-25 21:19:24追加了内容
哈，居然还有一个点！！！！
@黄依成
1
已采纳
中级天翼
这道题要用高精度的，酷町堂数据比较水
送你一个表：
string a[105]={"0","1","3","9","33","153","873","5913","46233","409113","4037913","43954713","522956313","6749977113","93928268313","1401602636313","22324392524313","378011820620313","6780385526348313","128425485935180313","2561327494111820313","53652269665821260313","1177652997443428940313","27029669736328405580313","647478071469567844940313","16158688114800553828940313","419450149241406189412940313","11308319599659758350180940313","316196664211373618851684940313","9157958657951075573395300940313","274410818470142134209703780940313","8497249472648064951935266660940313","271628086406341595119153278820940313","8954945705218228090637347680100940313","304187744744822368938255957323620940313","10637335711130967298604907294846820940313","382630662501032184766604355445682020940313","14146383753727377231082583937026584420940313","537169001220328488991089808037100875620940313","20935051082417771847631371547939998232420940313","836850334330315506193242641144055892504420940313","34289376947494122614363304694584807557656420940313","1439295494700374021157505910939096377494040420940313","61854558558074209658512637979453093884758552420940313","2720126133346522977702138448994068984204397080420940313","122342346998826717539665299944651784048588130840420940313","5624964506810915667389970728744906677010239883800420940313","264248206017979096310354325882356886646207872272920420940313","12678163798554051767172643373255731925167694226950680420940313","620960027832821612639424806694551108812720525606160920420940313","31035053229546199656252032972759319953190362094566672920420940313"};
至于怎么用，你肯定知道的
黄依成在2020-12-23 18:40:44追加了内容
啊这复制不了
0
高级光能
加上高精代码吧。。。
什么，你不会？
不然就用简洁的Python(自带高精yyds)
把你写的代码转换成Python的代码
然后就可以AC了
什么，你不会?
那我就无能为力了
毕竟我使用Python AC的
0
初级天翼
先写出递推关系式f[i]=f[i-1]*i，sum+=f[i];
再用高精度乘法和高精度加法替换这里乘法和加法，注意f[0]="1"
0
初级天翼
我还不知道怎么登洛谷
