hdu 5185 Equation（dp题）

最新推荐文章于 2019-09-09 11:53:00 发布

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本文探讨了一种特定的组合数学问题，即对于给定的整数n和m，寻找满足特定条件的组合数量，并通过模运算输出结果。该问题来源于BestCoderRound#32竞赛。

Equation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 647 Accepted Submission(s): 227

Problem Description
Gorwin is very interested in equations. Nowadays she gets an equation like this
x1+x2+x3+⋯+xn=n, and here
0≤xi≤nfor1≤i≤nxi≤xi+1≤xi+1for1≤i≤n−1

For a certain n, Gorwin wants to know how many combinations of xi satisfies above condition.
For the answer may be very large, you are expected output the result after it modular m.

Input
Multi test cases. The first line of the file is an integer T indicates the number of test cases.
In the next T lines, every line contain two integer n,m.

[Technical Specification]
1≤T<20
1≤n≤50000
1≤m≤1000000000

Output
For each case output should occupies one line, the output format is Case #id: ans, here id is the data number starting from 1, ans is the result you are expected to output.
See the samples for more details.

Sample Input

2
3 100
5 100

Sample Output

Case #1: 2
Case #2: 3

Source
BestCoder Round #32

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 50010
int dp[500][N];
int main()
{
    int T,t=1;
    long long n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld %lld",&n,&m);
        int k=1;
        while(k*(k+1)<=2*n) k++;
        k--;
        for(int i=0;i<=n;i++)
        dp[0][i]=0;
        for(int i=1;i<=k;i++)
            dp[i][0]=0;
        dp[0][0]=1;
        for(int i=1;i<=k;i++)
        {
            for(int j=i;j<=n;j++)
            {
                dp[i][j]=(dp[i][j-i]+dp[i-1][j-i])%m;
            }
        }
        long long ans=0;
        for(int i=1;i<=k;i++)
            ans=(ans+dp[i][n])%m;
        printf("Case #%d: %lld\n",t++,ans);
    }
    return 0;
}