HDU 1085 Ignatius and the Princess III (母函数-整数拆分)
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Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11780 Accepted Submission(s): 8340
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
41020
Sample Output
542627
如 4可以有5种方式:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
总共有5种,所以输出5
#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;const int MAX = 140;int ans[MAX],tmp[MAX];void work(int num){ int i,j,k; memset(ans,0,sizeof(ans)); memset(tmp,0,sizeof(tmp)); for(i=0;i<=num;++i){ ans[i] = 1; } //(author : CSDN iaccepted) for(i=2;i<=num;++i){ for(j=0;j<=num;++j){ for(k=0;k<=num && j+k*i<=num;++k){ tmp[j+k*i] += ans[j]; } } for(j=0;j<=num;++j){ ans[j] = tmp[j]; tmp[j] = 0; } }}int main(){ //freopen("in.txt","r",stdin); //(author : CSDN iaccepted) int num;work(120); while(scanf("%d",&num)!=EOF){ printf("%d\n",ans[num]); } return 0;}
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