P1009 [NOIP1998 普及组] 阶乘之和

解题思路

• 思路1：打表：因为n<50,所以可以借助python自带的高精度打表，再以字符串的形式存入数组中，见代码1（python联赛只能作为工具，不可以作为答案！）
• 思路2：老老实实用高精度加法+乘法，见代码2

解题代码1

f=open(r'point.txt','w')//python代码
sum=1
i=2
last=1
n=int(input())
while i<=n:
	last*=i
	sum+=last
	i+=1
	f.write('"')
	f.write(str(sum))
	f.write('",')
f.close	

#include<iostream>
using namespace std;
string a[60]={"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"};
int main()
{
	int n;
	cin>>n;
	cout<<a[n];
}

解题代码2

#include<bits/stdc++.h>
using namespace std;
int main(){
	int n,a[101]={0},b[101]={0};
	cin>>n;
	b[1]=a[1]=1;
	for(int i=2;i<=n;i++){
		for(int j=1;j<100;j++) b[j]=b[j]*i; 
		for(int j=1;j<100;j++){//进位处理 
			if(b[j]>9){
				b[j+1]+=b[j]/10;
				b[j]%=10;
			}
		}
		for(int j=1;j<100;j++) a[j]+=b[j];//将i的阶层加上去 
		for(int j=1;j<100;j++){//进位处理 
			if(a[j]>9){
				a[j+1]+=a[j]/10;
				a[j]%=10;
			}
		}
	}
	int k;
	for(k=100;k>=0&&a[k]==0;k--);
	for (int j=k;j>=1;j--) cout<<a[j];
	return 0;
}