NYOJ 420 P-Squared summation (fast Power + congruence theorem)

Title Description:

http://acm.nyist.net/JudgeOnline/problem.php?pid=420

A very simple question, beg 1^p+2^p+3^p+......+n^p and.

Input

The first line of a single number t represents the number of test data groups. Next there will be T-line numbers, each line consisting of two digital n,p,
Enter the guaranteed 0<n<=1000,0<=p<=1000.

Output

Output 1^p+2^p+3^p+......+n^p The result of 10003, each result is a single row.

Sample input

210 110 2

Sample output

55385

Topic Analysis:

The problem of fast Power + congruence theorem, although has been written many times the fast power but still do not remember, each time is to see the template, this time must remember it.

AC Code:

/** * Quick Power get touch + +/#include <iostream> #include <cstdio> #include <map> #include <cstring> #include <string> #include <algorithm> #include <queue> #include <vector> #include <stack># include<cstdlib> #include <cctype> #include <cstring> #include <cmath>using namespace Std;int MoD (int a,int b,int n) {    int t = 1;    if (b = = 0)        return 1;    if (b = = 1)         return a%n;    t = mod (A, b>>1, n);    t = t*t% n;    if (b&1) {//b is an odd        t = t*a% n;    }    return t; }int Main () {    int t;    cin>>t;    while (t--) {        int p,n;        cin>>n>>p;        int res=0;        for (int i=1;i<=n;i++) {            int t=mod (i,p,10003);            cout<<t<<endl;            Res= (res%10003+t%10003)%10003;        }        cout<<res<<endl;    } return 0;}        

NYOJ 420 P-Squared summation (fast Power + congruence theorem)