5185 Equation

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Equation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 656    Accepted Submission(s): 230

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 

题意：n个数满足  0<=X(i)<=n   X(i+1) 等于 X(i)或X(i)+1， 因此第一个X1 必定小于等于1 ( 若X1=2  则n个数最小为2*n ) 

          最理想的状态为1+2+3+...+x=n;  则可以求出xi最大值  [(1+x)*x]/2 =n;   

          其中x1~xn去重以后 一定是连续的 可以由dp[i][j]表示 由1~i种数字构成的和为j的组合数

          状态转移方程为dp[i][j]=dp[i][j-i]+dp[i-1][j-i]   ( dp[i][j-i]表示在n=j-i的组合数字的基础上在xn=i 后面增加一个数字i  ；dp[i-1][j-1] 在n=j-i的组合数字的基础上在xn=i-1 后面增加一个 数字i)

AC：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
typedef long long LL;
using namespace std;
int dp[500][50050]={0};
int main()
{
    int t;scanf("%d",&t);
    int v=1;
    while(t--)
    {
        LL n,mod;
        scanf("%lld%lld",&n,&mod);
        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=0;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])%mod;

        LL ans=0;
        for(int i=1;i<=k;i++)
            ans+=dp[i][n],ans%=mod;
        printf("Case #%d: %lld\n",v++,ans);

    }
}