Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26219 Accepted Submission(s): 18101
Problem Description
"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
Input
The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.
Output
For each test case, you have to output a line contains an integer P which indicate the different equations you have found.
Sample Input
4
10
20
10
20
Sample Output
5
42
627
42
627
题目大意:
求n有几种划分(3=1+2和3=2+1是同一种划分方案)。
dp[i][j]表示i分为j块一共有几种方案。
那么，一般的，考虑划分的j块中有多少个1，接着用截边法处理:
若有0个1，把这j块都减一，转化为i-j分为j块，dp[i][j]+=dp[i-j][j];
若有1个1，把这j块都减一，转化为i-j分为j-1块，dp[i][j]+=dp[i-j][j-1];
一直考虑到有k个1即可。
每个数的划分数即为sum dp[i][]。
#include <cstdio>
#include <cstring> using namespace std; const int maxn=; //动规打表
int dp[maxn+][maxn+]; int sum[maxn+]; int main()
{
memset(dp,,sizeof(dp));
memset(sum,,sizeof(sum));
for(int i=;i<=maxn;++i)
{
dp[i][]=dp[i][i]=;
for(int j=;j<=i-;++j)
{
for(int k=;k<=j;++k)
{
dp[i][j]+=dp[i-j][j-k];
}
}
for(int j=;j<=i;++j)
{
sum[i]+=dp[i][j];
}
} int n;
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",sum[n]);
} return ;
}