hdoj GCC 3123 （大数阶乘取余&转换）

GCC

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 4463    Accepted Submission(s): 1470


[align=left]Problem Description[/align]
The GNU Compiler Collection (usually shortened to GCC) is a compiler system produced by the GNU Project supporting various programming languages. But it doesn’t contains the math operator “!”.

In mathematics the symbol represents the factorial operation. The expression n! means "the product of the integers from 1 to n". For example, 4! (read four factorial) is 4 × 3 × 2 × 1 = 24. (0! is defined as 1, which is a neutral element in multiplication,
not multiplied by anything.)

We want you to help us with this formation: (0! + 1! + 2! + 3! + 4! + ... + n!)%m

 

[align=left]Input[/align]
The first line consists of an integer T, indicating the number of test cases.

Each test on a single consists of two integer n and m.

 

[align=left]Output[/align]
Output the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m.

Constrains

0 < T <= 20

0 <= n < 10^100 (without leading zero)

0 < m < 1000000

 

[align=left]Sample Input[/align]

1
10 861017

 

[align=left]Sample Output[/align]

593846
//由题意可知：当n>m时，从m-n这一区域对m取余都得0，所以不用计算。这样就简化了很多。此题有个坑点就是0!=1;这个必须得注意。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define ll long long
#define N 1000010
using namespace std;
char s[110];
ll sum
;
ll change(char *s)
{
int l,i,ans=0;
l=strlen(s);
for(i=0;i<l;i++)
ans=ans*10+(s[i]-'0');
return ans;
}
int main()
{
int t;
int n,m;
int i,j,k,l;
scanf("%d",&t);
while(t--)
{
scanf("%s%d",s,&m);
l=strlen(s);
if(l<7)
{
n=change(s);
if(n>m)
n=m-1;
sum[0]=1;
int ans=0;
for(i=1;i<=n;i++)
{
sum[i]=(sum[i-1]*i)%m;
ans=(ans+sum[i])%m;
}
printf("%d\n",(ans+1)%m);
}
else
{
n=m-1;
sum[0]=1;
int ans=0;
for(i=1;i<=n;i++)
{
sum[i]=(sum[i-1]*i)%m;
ans=(ans+sum[i])%m;
}
printf("%d\n",(ans+1)%m);
}
}
return 0;
}