Ignatius and the Princess III
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
41020
Sample Output
542627
f[i][j]表示把整数 i 拆成最多 j 个数字所具有的方法数。
那么if (i >ｊ) f[i][j] = f[i-j][j] + f[i][j-1]; 意思就是如果i>j，，那么有两种方式：一种是先把i里面分理处j个1，然后再把i-j拆成最多i-j个数字；另一种是把i拆分成最多j-1个数字。
if (i < j) f[i][j] = f[i][i]; 意思就是如果i<j，那么这种情况和把数字i最多拆成i个数字的是一样的。
if (i == j) f[i][j] = f[i][j-1] + 1; 意思就是如果i==j，那么可以把数字i拆分成j-1个数字，也可以把数字i拆分成i个1（这个就是那个1的意义）
另外，当i==1||j==1时，f[i][j]=1。
代码如下：
#include <cstdio>int main(){int n;while(scanf("%d",&n)==1){int i,j,f[125][125]={0};for(i=0;i<=n;i++)f[i][1]=f[1][i]=1;for(i=2;i<=n;i++)for(j=1;j<=n;j++){if(i>j) f[i][j]=f[i-j][j]+f[i][j-1];else if(i==j) f[i][j]=1+f[i][j-1];else f[i][j]=f[i][i];}printf("%d\n",f[n][n]);}return 0;}
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有勇气并不表示恐惧不存在，而是敢面对恐惧克服恐惧
