埃及分数 迭代加深搜索 IDA*

迭代加深搜索 IDA*

首先枚举当前选择的分数个数上限maxd，进行迭代加深

之后进行估价，假设当前分数之和为a,目标分数为b，当前考虑分数为1/c，那么如果1/c×（maxd - d）< a - b那么不可能得解，进行剪枝

#include <cstdio>
#include <cstring>
#include <algorithm>


const int maxn = 100;
const int inf = 2139062143;
int a, b;
int ans[maxn], cur[maxn];
int maxd = 0;


bool better(void) {
    for (int i = maxd; i >= 1; i--) {
        if (cur[i] < ans[i]) return (1);
        else if (cur[i] > ans[i]) return (0);
    }
}

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

int getfirst(int aa, int bb) {
    if (bb % aa == 0) return (bb / aa);
    else return (bb / aa + 1);
}

void dfs(int d, int aa, int bb) {
    if (d == maxd + 1) {
        if (!aa && better()) memcpy(ans, cur, sizeof(int) * (maxd + 1));
        return;
    }
    int fr = std :: max(getfirst(aa, bb), cur[d-1] + 1);
    for (; ; fr++) {
        if (aa * fr > (bb * (maxd - d + 1))) break;
        cur[d] = fr;
        int g = gcd(bb, fr);
        int ta = (aa * fr - bb) / g;
        int tb = (bb * fr / g);
        dfs(d + 1, ta, tb);
    }
}
int main () {
    scanf("%d %d", &a, &b);
    for (maxd = 1; ; maxd++) {
        memset(ans, 127, sizeof(ans));
        dfs(1, a, b);
        if (ans[1] > 0 && ans[1] != inf) {
            int cur = 1;
            while (ans[cur] > 0 && ans[cur] != inf) {
                printf("%d ", ans[cur]);
                cur++;
            }
            break;
        }
    }

    return 0;
}