uva 10910（子集和问题）

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Marks Distribution

In an examination one student appeared in N subjects and has got total T marks. He has passed in all the Nsubjects where minimum mark for passing in each subject is P. You have to calculate the number of ways the student can get the marks. For example, if N=3, T=34 and P=10 then the marks in the three subject could be as follows.

So there are 15 solutions. So F (3, 34, 10) = 15.

Input

In the first line of the input there will be a single positive integer K followed by K lines each containing a single test case. Each test case contains three positive integers denoting N, T and P respectively. The values of N, T and P will be at most 70. You may assume that the final answer will fit in a standard 32-bit integer.

Output

For each input, print in a line the value of F (N, T, P).

思路：过程的最后一步为前n门课总分为j的种类；

　　　　最后一步的子问题为：前n-1门课总分为j-k的种类；

　　　　显然；k>=p&&k<=j-p*(n-1);(保证每门课都能取到最低值)；

　　　　状态转移方程为：dp[i][j]+=dp[i-1][j-k];

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cstdlib>
 6 #include<iomanip>
 7 #include<cmath>
 8 #include<vector>
 9 #include<queue>
10 #include<stack>
11 using namespace std;
12 #define PI 3.141592653589792128462643383279502
13 int dp[75][75];
14 int n;
15 int main(){
16     //#ifdef CDZSC_June
17     //freopen("in.txt","r",stdin);
18     //#endif
19     //std::ios::sync_with_stdio(false);
20     cin>>n;
21     int N,T,P;
22     while(n--){
23         cin>>N>>T>>P;
24 
25         memset(dp,0,sizeof(dp));
26         for(int i=0;i<=70;i++)dp[1][i]=1;
27         for(int i=2;i<=N;i++)
28         for(int j=P;j<=T;j++){
29             for(int k=P;k<=j-P*(i-1);k++)
30             dp[i][j]+=dp[i-1][j-k];
31         }
32     cout<<dp[N][T]<<endl;
33     }
34     return 0;
35 }

uva 10910（子集和问题）

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原文地址：http://www.cnblogs.com/yoyo-sincerely/p/5095335.html