hdu_3123_GCC
2018-06-17 21:32:20来源:未知 阅读 ()
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
InputThe 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.
OutputOutput the answer of (0! + 1! + 2! + 3! + 4! + ... + n!)%m.
Constrains
0 < T <= 20
0 <= n < 10^100 (without leading zero)
0 < m < 1000000
Sample Input
1 10 861017
Sample Output
593846
唬人的题目,当x>=m, x! =0 mod m
我们就是在求小于m的 ∑x!
提取公因式化简得
1+(1*2)+(1*2*3)+.....+(1*2*3*....*x)=1*(1+2*(1+3*(.....x-1*(1+x))))
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<algorithm> using namespace std; int main() { char n[105]; int m,t; scanf("%d",&t); while(t--) { scanf("%s",n); getchar(); scanf("%d",&m); int s=strlen(n); int n1=0; for(int i=0;i<s;i++) { n1*=10; n1+=n[i]-'0'; if(n1>=m) break; } long long ans; if(n1>=m) { ans=m-1; for(int i=m-2;i>=1;i--) { ans=(ans+1)%m*i%m; } } else { ans=n1; for(int i=n1-1;i>=1;i--) { ans=(ans+1)%m*i%m; } } ans++; cout<<ans%m<<endl; } }
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