题目大意
思路分析
给一个递归函数,直接施展用时太长了,需要将中间结果保留下来。虽然题比较水,但是增强了自己对递归的理解,无需在每一个if条件中判断各个因子是否存在,如果他存在下一次递归就会直接返回,如果不存在,下一次递归也会将他找到并赋值。
#include<iostream>
#include<cmath>
#include<string.h>
#include<string>
#include<algorithm>
#include<vector>
#include<map>
#include<cstdio>
using namespace std;
#define MAX 25
#define inf 1e10
#define ll int
ll v[MAX][MAX][MAX];//有任意大于20的数都会回到20
ll w(ll a, ll b, ll c) {
ll res = 0;
if (a <= 0 || b <= 0 || c <= 0)return 1;
if (a > 20 || b > 20 || c > 20) return w(20, 20, 20);
if (v[a][b][c] != -1)return v[a][b][c];
if (a < b&&b < c) {
return v[a][b][c] = w(a, b, c - 1) + w(a, b - 1, c - 1) - w(a, b - 1, c);
}
else {
return v[a][b][c] = w(a - 1, b, c) + w(a - 1, b - 1, c) + w(a - 1, b, c - 1) - w(a - 1, b - 1, c - 1);
}
}
int main() {
int a, b, c;
memset(v, -1, sizeof(v));
while (scanf("%d%d%d", &a, &b, &c) && (a != -1 || b != -1 || c != -1)) {
printf("w(%d, %d, %d) = %d\n", a, b, c, w(a, b, c));
}
}