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Ignatius and the Princess III(母函数一种技巧性的暴力)
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13070 Accepted Submission(s): 9236
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
题意:整数划分,给你一个整数n,问有多少种方式划分整数;比如2可以划分为2,1+1;
意解:母函数模板; 母函数其实就是把组合加法与幂指数的乘法法则联合在一起
#include <iostream> #include <cstring> #include <cstdio> using namespace std; const int M = 1e4 * 2; typedef long long ll; ll a[M],b[M]; /********/ /* * 母函数,a数组是维护当前能组成的数的方法数,b数组为中间量; * 母函数其实就是把组合加法于幂指数的乘法法则联合在一起 * 这相当于暴力查找符合题意的方法种数,取于不取的问题,取几个的问题; */ int main() { int n; while(~scanf("%d",&n)) { for(int i = 0; i <= n; i++) { a[i] = 1; b[i] = 0; } for(int i = 2; i <= n; i++) { for(int j = 0; j <= n; j++) //当前序列的指数 { for(int k = 0; k + j <= n; k += i) //接下来序列的指数,因为指数是以i递增的; { b[k + j] += a[j]; } } for(int j = 0; j <= n; j++) //更新序列; { a[j] = b[j]; b[j] = 0; } } printf("%I64d\n",a[n]); } return 0; }
Ignatius and the Princess III(母函数一种技巧性的暴力)
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