hdi2189基础dp

/*****************************************
Author      :Crazy_AC(JamesQi)
Time        :2015
File Name   :
思路:题意就是又不超过n个素数组合成n的情况有多少种;
1 2 3 4 5 6 7 8 9 。。。可以看成是数的生成情况;
就是dp[i][j]表示又不超过i的素数组合成j的情况总和;
那么:1.i是素数时dp[i][j] += dp[i][j - i];
2.dp[i][j] = dp[i - 1][j];
*****************************************/
// #pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <sstream>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
using namespace std;
#define MEM(a,b) memset(a,b,sizeof a)
#define pk push_back
template<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}
template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}
typedef long long ll;
typedef pair<int,int> ii;
const int inf = 1 << 30;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int N = 200;
bool prime[N + 1];
int dp[N][N];
inline void Init(){
	MEM(prime, false);
	// prime[1] = false;
	// prime[0] = true;
	for (int i = 2;i <= 200;++i){
		if (!prime[i]){
			for (int j = i + i;j <= N;j += i){
				prime[j] = true;
			}
		}
	}
}
int main()
{	
	// ios::sync_with_stdio(false);
	// freopen("in.txt","r",stdin);
	// freopen("out.txt","w",stdout);
	int n,T;
	Init();
	cin >> T;
	while(T--){
		cin >> n;
		MEM(dp, 0);
		dp[1][0] = 1;
		for (int i = 2;i <= n;++i){
			for (int j = 0;j <= n;++j)
				dp[i][j] = dp[i - 1][j];
			if (prime[i]) continue;
			for (int j = i;j <= n;++j)
				dp[i][j] += dp[i][j - i];
		}
		cout << dp[n][n] << '\n';
	}
	return 0;
}

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