HDU 3123 GCC(數學)

GCC

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 3190    Accepted Submission(s): 1004


Problem DescriptionThe 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 

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

110 861017

 

Sample Output

593846

 

Source2009 Asia Wuhan Regional Contest Online 

Recommendlcy 這個題目又悲催了幾次，命名按照我的計算是不會逾時的，但是交上去老是逾時，原來是longlong的問題，哎後來想改成__int64無奈linux下G++不支援__int64我也沒管直接在代碼上替換，然後提交，果然0msAC這個數學題目其實還是比較簡單的分析如下：1.這個題目給的資料很嚇人，10^100還是階乘，難道還用大數模數？ 都不用，因為後面說了對N!%M,那麼M的最大為一百萬那麼如若N>M N直接就等於M，後面全部是02.第二是不是每次要求新的階乘呢？ 當然不是，在上一次的基礎上迭代一下就OK 了，所以這個題目就沒什麼難度了！

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int main(){int t;scanf("%d",&t);__int64 Max;__int64 ans;__int64 futher,num,mod;__int64 i,j,k;char str[200];while(t--){scanf("%s",str);cin>>mod;if(strlen(str) >= 7)num=1000000;elsesscanf(str,"%I64d",&num);if(num > mod)num=mod;futher=1;ans=1;for(i=1;i<=num;i++){futher=(futher*i)%mod;if(futher == 0)break;ans+=futher;}cout<<ans%mod<<endl;}return 0;}