HDU-1028 Ignatius and the Princess III(生成函数)

题意

给出$n$，问用$1$到$n$的数字问能构成$n$的方案数

思路

生成函数基础题，$x^{n}$的系数即答案。

代码

#include 
#define DBG(x) cerr << #x << " = " << x << endl;

using namespace std;

const int N = 120 + 5;

int n, c[2][N];

int main() {
    while(~scanf("%d", &n)) {
        for(int i = 0; i <= n; i++) c[1][i] = 1, c[0][i] = 0;
        for(int i = 2; i <= n; i++) {
            for(int j = 0; j <= n; j++) {
                for(int k = 0; k + j <= n; k += i) c[i & 1][k + j] += c[1 - (i & 1)][j];
            }
            for(int j = 0; j <= n; j++) c[1 - (i & 1)][j] = 0;
        }
        printf("%d\n", c[n & 1][n]);
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/DuskOB/p/10604288.html