题目链接

Equation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 49    Accepted Submission(s): 18

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

23 1005 100

 

Sample Output

Case #1: 2Case #2: 3

/*这可以看成是一个完全背包问题，但是由于数字比较多直接DP会超时。其实可以发现数字的种类最多是sqrt(n)级别的。那么就可以把复杂度降到nsqrt(n)；Dp[i][j]代表前i个数字放好之后能组成j的和数是多少。递推方程是Dp[i][j]=dp[i][j-i]+dp[i-1][j-i]; 表示第i种数字放的时候，前面要么放了i，要么放了i-1边界条件是dp[i][j]=0 for i < jDp[0][0]=1;所以最后总的复杂度是n*sqrt(n)*/#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;int n,M,tt;int dp[50005][305];int main() {    //freopen("D://imput.txt","r",stdin);    int cur=1;    scanf("%d",&tt);    while(tt--) {        scanf("%d%d",&n,&M);        for(int i=1; i<=n; i++)            for(int j=1; j*(j+1)/2<=i; j++)                dp[i][j]=0;        dp[1][1]=1;        for(int i=1; i<=n; i++) {            for(int j=1; j*(1+j)/2<=i&&i+j<=n; j++) {                dp[i+j][j]=(dp[i+j][j]+dp[i][j])%M;                if(i+j+1<=n)                    dp[i+j+1][j+1]=(dp[i+j+1][j+1]+dp[i][j])%M;            }        }        int res=0;        for(int i=1; i*(i+1)/2<=n; i++) {            res=(res+dp[n][i])%M;        }        printf("Case #%d: %d\n",cur++,res);    }    return 0;}