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!"
"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
Sample Output
5 42 627
题目大意:每个数可以有几个组合,给你一个数请你输出可以组合个个数,3+1和1+3是同样的。
思路:母函数。
代码:
#include<stdio.h> #include<string.h> #define max 120 int a[max+1]; //数组a代表最后的结果 int b[max+1]; //数组b代表中间值 int main() { int i,j,k,N; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); a[0]=1; for(i=1;i<=max;i++) { for(j=0;j*i<=max;j++) { for(k=0;k+j*i<=max;k++) { b[k+j*i]+=a[k]; } } for(j=0;j<=max;j++) { a[j]=b[j]; } memset(b,0,sizeof(b)); } while(scanf("%d",&N)!=EOF) { printf("%d\n",a[N]); } return 0; }
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