Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24796 Accepted Submission(s): 17138
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
4 10 20
Sample Output
5 42 627
Author
Ignatius.L
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生成函数的裸题
对于每个数构造一个多项式$ax^i$表示权值为$i$的方案数为$a$
初始时$a$为1
然后全乘起来就行
#include<cstdio>
#include<cstring>
#include<algorithm>
const int MAXN = 121;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int cur[MAXN], nxt[MAXN];
main() {
int N;
while(scanf("%d", &N) != EOF) {
memset(cur, 0, sizeof(cur));
for(int i = 0; i <= N; i++) cur[i] = 1;
for(int i = 2; i <= N; i++) {
for(int j = 0; j <= N; j++)
for(int k = 0; j + i * k <= N; k++)
nxt[j + k * i] += cur[j];
memcpy(cur, nxt, sizeof(nxt));
memset(nxt, 0, sizeof(nxt));
}
printf("%d\n", cur[N]);
}
return 0;
}