38 lines
791 B
C++
38 lines
791 B
C++
#include <bits/stdc++.h>
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using namespace std;
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const int INF = 0x3f3f3f3f;
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const int N = 15;
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int a[N], b[N], n;
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int ans = INF;
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/**
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* @param step 第几步
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* @param x 目前的酸度值
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* @param y 目前的苦度值
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*/
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void dfs(int step, int x, int y) {
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if (step == n) {
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//清水不行~
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if (x == 0 && y == 0) return;
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//更新ans
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ans = min(abs(x - y), ans);
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return;
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}
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//选择
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dfs(step + 1, (x == 0 ? 1 : x) * a[step + 1], y + b[step + 1]);
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//放弃
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dfs(step + 1, x, y);
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}
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int main() {
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//n种配料
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cin >> n;
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//分别读入酸度和苦度
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for (int i = 1; i <= n; i++) cin >> a[i] >> b[i];
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dfs(0, 0, 0);
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//输出
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printf("%d\n", ans);
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return 0;
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} |