标签：his end it! res code fine rmi include iss
Ignatius and the Princess III
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25867 Accepted Submission(s): 17879
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
Sample Output
G(x) = (1 + x^1 + x^2 + x^3 ......) * (1 + x^2 + x^4 + x^8.....) * (1 + x^3 + x^6 + x^9 ......)...........
x^k的系数即为数字k的拆分方案数
1 #include <bits/stdc++.h>
2 using namespace std;
3 int c1[125],c2[125];
4 int n;
5
6 int build(){
7 for (int i = 0;i <= 120;++i){
8 c1[i] = 1,c2[i] = 0;
9 }
10 for (int i = 2;i <= 120;++i){
11 if (c1[i] == 0) return 0 * printf("%d fuck",i);
12 for (int j = 0;j <= 120;++j){
13 for (int k = 0;k+j <= 120;k+=i){
14 c2[k+j] += c1[j];
15 }
16 }
17 for (int k = 0;k <= 120;++k){
18 c1[k] = c2[k];
19 c2[k] = 0;
20 }
21 }
22 }
23
24
25 int main(){
26 build();
27 while(~scanf("%d",&n)){
28 printf("%d\n",c1[n]);
29 }
30 }
HDU1028 Ignatius and the Princess III 母函数
标签：his end it! res code fine rmi include iss
原文地址：https://www.cnblogs.com/mizersy/p/9532053.html
