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Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3514 Accepted Submission(s): 1998
dp[i][j]表示用前i种水果弄到j个水果的组合数
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <vector> #include <queue> #include <map> #include <set> #include <stack> #include <algorithm> using namespace std; #define root 1,n,1 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define lr rt<<1 #define rr rt<<1|1 typedef long long LL; typedef pair<int,int>pii; #define X first #define Y second const int oo = 1e9+7; const double PI = acos(-1.0); const double eps = 1e-6 ; const int N = 305 ; const int mod = 1e9+7; LL dp[N][N]; int n , m , a[N] , b[N] ; void init() { memset( dp , 0 , sizeof dp ); dp[0][0] = 1 ; for( int i = 1 ; i <= n ; ++i ) { for( int j = 0 ; j <= m ; ++j ){ for( int k = a[i] ; k <= b[i] ; ++k ){ if( j < k ) continue ; dp[i][j] += dp[i-1][j-k]; } } } } void Run() { for( int i = 1 ; i <= n ; ++i ){ cin >> a[i] >> b[i] ; } init(); cout << dp[n][m] << endl ; } int main() { //freopen("in.txt","r",stdin); ios::sync_with_stdio(false); while( cin >> n >> m ) Run(); }
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原文地址:http://www.cnblogs.com/hlmark/p/4204043.html