hdu 5185 Equation（dp题）

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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hujie

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 50010
int dp[500]
;
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]
)%m;
printf("Case #%d: %lld\n",t++,ans);
}
return 0;
}