D. A Simple But Difficult Problem

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D. A Simple But Difficult Problem

Time Limit: 5000ms

Memory Limit: 65536KB

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

SubmitStatus PID: 4356

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 1100 2 -1 -1

Sample Output

05050

38350

看的别人的代码

#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,Rconst 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\n",sum);    }    return 0;}

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