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Question 192. Luogu P1009 Simulation-[NOIP1998 Popularization Group] Sum of Factorials
1. The topic
2. Problem solving
Because the n of this question can be very large, the result after calculating the factorial sum cannot be expressed and output, so it is necessary to turn the string to simulate the calculation process - the shell remains unchanged, the factorial is still recursively calculated, and the loop summation is used, which is the key The summation and multiplication of the effects are solved by analog vertical calculations. code show as below:
#include <bits/stdc++.h>
using namespace std;
string Add(string a,string b)
{
string ans;
int add=0;//需要加到本位的进位数
int i,j;
for(i=a.length()-1,j=b.length()-1;i>=0&&j>=0;i--,j--)
{
int res=(a[i]-'0')+(b[j]-'0')+add;//和等于两数之和加上进位
ans=to_string(res%10)+ans;//只将和的个位数存入结果
add=res/10;//计算进位
}
while(i>=0)
{
int res=(a[i--]-'0')+add;
ans=to_string(res%10)+ans;
add=res/10;
}
while(j>=0)
{
int res=(b[j--]-'0')+add;
ans=to_string(res%10)+ans;
add=res/10;
}
if(add!=0)//如果进位不是0就还得算上
{
ans=to_string(add)+ans;
}
return ans;
}
string Multiple(string a,string b)
{
string tmp[2001];//存储列竖式计算乘法时的中间结果
string ans="0";
if(a[0]==0||b[0]==0)
{
ans="0";
return ans;
}
int alen=a.length(),blen=b.length();
for(int i=blen-1; i>=0; i--)
{
int preadd=0,add=0;//本位的进位，下一位的进位
for(int j=0; j<(blen-1)-i; j++)
{
tmp[i]="0"+tmp[i];//用0补位,便于后序计算
}
for(int j=alen-1; j>=0; j--)
{
int res=(a[j]-'0')*(b[i]-'0');
add+=res/10;//得到下一位进位
res=res%10;//本位结果只留个位数
res+=preadd;//本位结果还需加上本位的进位
add+=res/10;//本位结果又做了加法所以得再更新一回下一位进位
res=res%10;
preadd=add;//更新本位的进位，给下一轮使用
add=0;//下一位进位再初始化为0
tmp[i]=to_string(res)+tmp[i];
}
if(preadd!=0)//如果进位不是0就还得算上
{
tmp[i]=to_string(preadd)+tmp[i];
}
ans=Add(ans,tmp[i]);
}
return ans;
}
string getFactorial(int n)
{
if(n==1)
{
return "1";//原本为1
}
else
{
return Multiple(to_string(n),getFactorial(n-1));//原本为n*getF...(n-1)
}
}
int main()
{
int n;
cin>>n;
string res="0";
for(int i=1;i<=n;i++)
{
res=Add(res,getFactorial(i));
}
cout<<res<<endl;
}