5185 Equation

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);

    }
}