LightOJ1213 Fantasy of a Summation
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题目链接:点我
If you think codes, eat codes then sometimes you may get stressed. In your dreams you may see huge codes, as I have seen once. Here is the code I saw in my dream.#include <stdio.h>int cases, caseno;int n, K, MOD;int A[1001];int main() { scanf("%d", &cases); while( cases-- ) { scanf("%d %d %d", &n, &K, &MOD); int i, i1, i2, i3, ... , iK; for( i = 0; i < n; i++ ) scanf("%d", &A[i]); int res = 0; for( i1 = 0; i1 < n; i1++ ) { for( i2 = 0; i2 < n; i2++ ) { for( i3 = 0; i3 < n; i3++ ) { ... for( iK = 0; iK < n; iK++ ) { res = ( res + A[i1] + A[i2] + ... + A[iK] ) % MOD; } ... } } } printf("Case %d: %d\n", ++caseno, res); } return 0;}Actually the code was about: 'You are given three integers n, K, MOD and n integers: A0, A1, A2 ... An-1, you have to write K nested loops and calculate the summation of all Ai where i is the value of any nested loop variable.'
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.Each case starts with three integers: n (1 ≤ n ≤ 1000), K (1 ≤ K < 231), MOD (1 ≤ MOD ≤ 35000). The next line contains n non-negative integers denoting A0, A1, A2 ... An-1. Each of these integers will be fit into a 32 bit signed integer.
Output
For each case, print the case number and result of the code.
Sample Input
23 1 350001 2 32 3 350001 2
Sample Output
Case 1: 6Case 2: 36
题意:
求和
思路:
这里共进行了
代码:
#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<iostream>using namespace std;typedef long long LL;LL qpow(LL x,LL n, LL mod){ LL ans = 1; while(n){ if(n&1) ans =(ans * x) % mod; x = (x *x) % mod; n >>= 1; } return ans;}int main(){ LL n,k,mod; int t,kase = 0; scanf("%d", &t); while( t--){ scanf("%lld %lld %lld",&n, &k,&mod); LL sum = 0; for(int i = 1; i <= n; ++i){ LL x; scanf("%lld", &x); sum += x; } sum %= mod; printf("Case %d: %lld\n", ++kase,sum* k*qpow(n, k - 1, mod) % mod); } return 0;}
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