ACdream 1007 （快速幂）

题目链接

a + b

Time Limit: 2000/1000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

Problem Description

Input

Output

Sample Input

2
3 1
1 2 3
3 10
1 2 3

Sample Output

6
60074

/************************************************************************* > File Name: 1007.cpp > Author: Stomach_ache > Mail: [email protected] > Created Time: 2014年08月11日 星期一 07时44分39秒 > Propose: ************************************************************************/ #include <cstdio>
/*Let's fight!!!*/ typedef long long LL; const LL mod = (LL)1e10 + 7; LL mul(LL a, LL b) { LL res = 0; while (b) { if (b&1) if ((res += a) >= mod) res -= mod; a <<= 1; if (a >= mod) a -= mod; b >>= 1; } return res; } LL mod_pow(LL a, LL b) { LL res = 1; while (b) { if (b&1) res = mul(res, a); a = mul(a, a); b >>= 1; } return res; } int main(void) { int t; scanf("%d", &t); while (t--) { LL n, k, ans = 0, x; scanf("%lld %lld", &n, &k); k %= 9560995488LL; for (int i = 0; i < n; i++) { scanf("%lld", &x); ans += mod_pow((x%mod+mod)%mod, k); if (ans >= mod) ans -= mod; } printf("%lld\n", ans); } return 0; }