BNU 4356 ——A Simple But Difficult Problem——————【快速幂、模运算】

A Simple But Difficult Problem

Time Limit: 5000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

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 计算前n个正整数的k次幂之和：

结果可能会非常非常大，输出结果的最后5位即可。

 

Input

输入数据有多组。每组数据一行，每行包括两个整数n和k (1<=n<=1,000,000,000,1<=k<=1000)。

输入数据以-1 -1 结束。

 

Output

对每一组输入数据，输出单独一行，包括一个整数，给出对应的答案。

 

Sample Input

100 1
100 2 
-1 -1

Sample Output

05050
38350

Source

第九届北京师范大学程序设计竞赛决赛

 

 

解题思路：快速幂解决超时问题。

 

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<string>
#include<iostream>
#include<queue>
#include<vector>
#include<set>
using namespace std;
typedef long long LL;
#define mid (L+R)/2
#define lson rt*2,L,mid
#define rson rt*2+1,mid+1,R
const int INF = 0x3f3f3f3f;
const int maxn = 1e5 + 300;
const int mod = 1e5;
LL qpowmod(LL n,LL k){
    LL ret = 1;
    while(k){
        if(k&1)
            ret = (ret*n) % mod;
        k = k>>1;
        n = n*n % mod;
    }
    return ret;
}
int main(){
    LL n, k;
    while(scanf("%lld%lld",&n,&k)!=EOF){
        if(n==-1 && k==-1) break;
        LL sum = 0;
        LL mo = n%mod, quotient = n/mod;
        if(quotient){
            for(LL i = 1;i <= mod; i++){
                sum = (sum + qpowmod(i,k)) % mod;
            }
            sum = (sum*quotient) % mod;
        }
        for(LL i = 1; i <= mo; i++){
            sum = (sum + qpowmod(i,k))%mod;
        }
        printf("%05lld
",sum);
    }
    return 0;
}