"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!"
InputThe 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.
OutputFor 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 代码：
/* gyt
Live up to every day */
#include<cstdio>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>`
#include<queue>
#include<set>
#include<string>
#include<map>
#include <time.h>
#define PI acos(-1)
using namespace std;
typedef long long ll;
typedef double db;
const int maxn = 10000+5;
const ll maxm = 1e7;
const ll mod = 1e9 + 7;
const int INF = 0x3f3f3f;
const ll inf = 1e15 + 5;
const db eps = 1e-9;
int c1[maxn], c2[maxn];
void solve() {
int n;
while(scanf("%d", &n)!=EOF) {
for (int i=0; i<=n; i++) {
c1[i]=1, c2[i]=0;
}
for (int i=2; i<=n; i++) {
for (int j=0; j<=n; j++) {
for (int k=0; j+k<=n; k+=i) {
c2[j+k]+=c1[j];
}
}
for (int j=0; j<=n; j++) {
c1[j]=c2[j];
c2[j]=0;
}
}
printf("%d\n", c1[n]);
}
}
int main() {
int t = 1;
//freopen("in.txt", "r", stdin);
//scanf("%d", &t);
while(t--)
solve();
return 0;
}