POJ 1980 DFS 剪枝

简单dfs，dfs(now,son,mother,cnt,res)，其中now表示当前可以用的分母值，son和mother表示当前分数值的分子与分母，cnt表示当前分分数个数，res表示当前分数的分母乘积，注意几步剪枝
1.分母乘积大于a（等于也可能符合条件不能剪）
2.分数的个数已经大于等于n
3.当前分数值已经大于目标分数值
Code

750ms

#include <stdio.h>
int p, q, n, a, ans;
void dfs(int now, int son, int mother, int cnt, int res){
if (p * mother == q * son && mother){
ans++; return;
}
if (res > a) return;
if (cnt >= n) return;
if (p * mother < q * son) return;
int tson, tmother;
if (mother) tson = son * now + mother, tmother = mother * now;
else tson = 1, tmother = now;
if (res * now <= a) dfs(now, tson, tmother, cnt + 1, res * now);
if (res * (now + 1) <= a) dfs(now + 1, son, mother, cnt, res);
}
int main(){
while (~scanf("%d%d%d%d", &p, &q, &a, &n), p || q || n || a){
ans = 0;
dfs(1, 0, 0, 0, 1);
printf("%d\n", ans);
}
return 0;
}

370ms

#include <iostream>
using namespace std;
int p, q, a, n, ans;

int gcd(int x, int y){
return x % y == 0 ? y : gcd(y, x % y);
}

int lcm(int x, int y){
return x / gcd(x, y) * y;
}

void dfs(int cur, int pre, int up, int down, int product){
for (int i = pre; i * product <= a; ++i){
int k = lcm(down, i);
int t1 = k / down * up;
int t2 = k / q * p;
if (t2 > t1 + k / i * (n - cur + 1))
return;
if (t2 == t1 + k / i)
++ans;
else if (t2 > t1 + k / i && cur < n)
dfs(cur + 1, i, t1 + k / i, k, product * i);
}
}

int main(){
while (scanf("%d%d%d%d", &p, &q, &a, &n) == 4){
if (p == 0 && q == 0 && a == 0 && n == 0)
break;
ans = 0;
dfs(1, 1, 0, q, 1);
printf("%d\n", ans);
}
return 0;
}

暴力

975ms....

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

#define LL long long
int const MAX = 1e6 + 1;
int const INF = 1 << 30;
double const EPS = 0.00000001;
using namespace std;

int p, q, n, t, a, ans;

//暴力
//cnt第几个分数,当前分母cur,当前和up/down,当前分母乘积mul
void dfs(int cnt, int cur, int up, int down, int mul){
// printf("%d %d %d %d %d\n", cnt, cur, up, down, mul);
if (cnt > n || mul > a || up * q > p * down) return;

if (p * down == q * up){
ans++; return;
}

//选择当前分数
dfs(cnt + 1, cur, up * cur + down, down * cur, mul * cur);

//不选当前分数
if (cur * mul <= a)
dfs(cnt, cur + 1, up, down, mul);

}
int main(){
while (scanf("%d%d%d%d", &p, &q, &a, &n), q){
ans = 0;
dfs(0, 1, 0, 1, 1);
printf("%d\n", ans);
}
return 0;
}

剪枝

47ms

减掉当前最大值也不能达到的点

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>

#define LL long long
int const MAX = 1e6 + 1;
int const INF = 1 << 30;
double const EPS = 0.00000001;
using namespace std;

int gcd(int a, int b){
return b ? gcd(b, a % b) : a;
}
int p, q, n, t, a, ans;

//暴力
//cnt已经选择几个分数,当前分母cur,当前和up/down,当前分母乘积mul
void dfs(int cnt, int cur, int up, int down, int mul){
// printf("%d %d %d %d %d\n", cnt, cur, up, down, mul);
if (cnt > n || mul > a || up * q > p * down) return;

if ((q * down * (n - cnt) + up * q * cur) < p * down * cur)
return;

if (p * down == q * up){
ans++; return;
}
//选择当前分数
dfs(cnt + 1, cur, up * cur + down, down * cur, mul * cur);

//不选当前分数
if (cur * mul <= a)
dfs(cnt, cur + 1, up, down, mul);

}
int main(){
while (scanf("%d%d%d%d", &p, &q, &a, &n), q){
ans = 0;
dfs(0, 1, 0, 1, 1);
printf("%d\n", ans);
}
return 0;
}