Fantasy of a Summation
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
", ++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.
OutputFor each case, print the case number and result of the code.
Sample Input2
3 1 35000
1 2 3
2 3 35000
1 2
Sample OutputCase 1: 6
Case 2: 36
分析:代码中k个循环执行加法的所有次数为k*n^k次,由k个循环的对称性可知,
数组中每个数被加的次数为k*n^(k-1)次,因此答案就是∑(a[i]*k*n^(k-1)),0<i<n,
即sum*k*n^(k-1),然后用快速幂做。
#include<cstdio> long long N,K,mod; long long pow(long long x,long long k) { long long res=1; while(k>0) { if(k&1) res=res*x%mod; k>>=1; x=x*x%mod; } return res; } int main() { int T,cas=0; scanf("%d",&T); while(T--) { long long sum=0,x; scanf("%lld%lld%lld",&N,&K,&mod); for(int i=0;i<N;i++) { scanf("%lld",&x); sum+=x; } sum%=mod; long long ans=pow(N,K-1); printf("Case %d: %lld ",++cas,ans%mod*K%mod*sum%mod); } return 0; }