LightOJ 1213 - Fantasy of a Summation (排列组合+快速幂)

1213 - Fantasy of a Summation

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

If you think codes, eat codes then sometimes you may getstressed. In your dreams you may see huge codes, as I have seen once. Here isthe 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... A_n-1, you have to write K nested loops and calculatethe summation of all A_i where i is the value of anynested 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 ... A_n-1. Each of these integerswill be fit into a 32 bit signed integer.

Output

For each case, print the case number and result of the code.

Sample Input	Output for Sample Input
2 3 1 35000 1 2 3 2 3 35000 1 2	Case 1: 6 Case 2: 36

 

题意：根据代码看就可以了

思路：由代码可得整个过程总共会加n^k次，但是每个数并不是加了n^k次，手写一下，用排列组合算下应该是n^(k-1)*k次，所以说ans=sum*n^(k-1)*k,，中间过程中取模就好了。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#define MAXN 1010000
#define LL long long
#define ll __int64
#include<iostream>
#include<algorithm>
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
LL gcd(LL a,LL b){return b?gcd(b,a%b):a;}
LL lcm(LL a,LL b){return a/gcd(a,b)*b;}
ll powmod(ll a,LL b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
int main()
{
	int t;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		int n,k,mod;
		scanf("%d%d%d",&n,&k,&mod);
		ll ans=0;
		for(int i=0;i<n;i++)
		{
			int a;
			scanf("%d",&a);
			ans=(ans+a)%mod;
		}
		ll num=powmod(n,k-1,mod);
		printf("Case %d: ",++cas);
		printf("%d\n",(ans*num*k)%mod);
	}
	return 0;
}