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http://lightoj.com/volume_showproblem.php?problem=1213
Description
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
2
3 1 35000
1 2 3
2 3 35000
1 2
Sample Output
Case 1: 6
Case 2: 36
#include<stdio.h> #include<math.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; const int N = 1010; const int INF = 0x3f3f3f3f; typedef long long ll; int mod; ll Pow(int a, int b, int c) { ll ans = 1; a %= c; while(b) { if(b % 2 == 1) ans = (ans * a) % c; a = (a * a) % c; b /= 2; } return ans; } int main() { int t, a[N], p = 0;; int n, k; ll sum; scanf("%d", &t); while(t--) { p++; sum = 0; scanf("%d%d%d", &n, &k, &mod); for(int i = 0 ; i < n ; i++) { scanf("%d", &a[i]); sum += a[i]; } ll s; s = Pow(n, k - 1, mod); s *= k; sum %= mod; sum *= s; sum %= mod; printf("Case %d: %lld\n", p, sum); } return 0; } /* 3 2 4 3 1 30 4 9 5 22 18 2 22 2 2147483647 3333 2147483647 2147483647 */
LightOJ 1213 Fantasy of a Summation(规律 + 快数幂)
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原文地址：http://www.cnblogs.com/qq2424260747/p/4942627.html
