标签：i++ open integer each view set 公式 ext ref
题目链接：http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1556
Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on the door. The type of the sequence of number is
But Jerry’s mathematics is poor, help him to escape from the room.
There are some cases (about 500). For each case, there are two integer numbers n, m describe as above ( 1 <= n < 1 000 000, 1 <= m < 1000).
For each case, you program will output the answer of the sum of the sequence of number (mod 1e9+7).
4 1 5 1 4 2 5 2 4 3
10 15 30 55 100
看sample猜公式。QAQ
公式是 1^m + 2^m + ···· + n^m。用快速幂解决。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 #define pb push_back 14 #define mp make_pair 15 #define ms(a, b) memset((a), (b), sizeof(a)) 16 //#define LOCAL 17 #define eps 0.0000001 18 #define LNF (1<<60) 19 typedef long long LL; 20 const int inf = 0x3f3f3f3f; 21 const int maxn = 100000+10; 22 const int mod = 1e9+7; 23 24 LL qpow(LL a, LL b) 25 { 26 LL ans = 1; 27 LL base = a; 28 while(b){ 29 if(b&1) ans = (ans * base)%mod; 30 b >>= 1; 31 base = (base * base)%mod; 32 } 33 return ans; 34 } 35 int main() 36 { 37 #ifdef LOCAL 38 freopen("input.txt", "r", stdin); 39 // freopen("output.txt", "w", stdout); 40 #endif // LOCAL 41 42 int n,m; 43 while(~scanf("%d%d", &n, &m)) 44 { 45 LL ans = 0; 46 for(int i = 1;i<=n;i++) 47 { 48 ans = (ans + qpow(i, m))%mod; 49 } 50 printf("%lld\n", ans); 51 } 52 return 0; 53 }
标签：i++ open integer each view set 公式 ext ref
原文地址：http://www.cnblogs.com/denghaiquan/p/6745244.html
