C Super Ryuma

Description

Solution

观察可得最多移动三次即可从起点到终点。
分情况讨论：

起点和终点重合，总步数为 0
共对角线或者曼哈顿距离不超过 3，总步数为 1
走两次对角线，即与起点曼哈顿距离为偶数，总步数为 2
起点的近邻点，即与起点曼哈顿距离不超过 6，总步数为 2
与对角线的距离曼哈顿距离不超过 3，总步数为 2
其它情况，总步数为 3

Code

Super Ryuma

#include <bits/stdc++.h>
#define int long long

using namespace std;

int a, b, c, d;

signed main() {
    cin >> a >> b >> c >> d;
    if (a == c && b == d) cout << 0 << endl;
    else if (a + b == c + d || a - b == c - d || (abs(a - c) + abs(b - d) <= 3)) {
        cout << 1 << endl;
    } else if ((abs(a - c) + abs(b - d)) & 1){
        if (abs(a + b - c - d) <= 3) cout << 2 << endl;
        else if (abs(a - b - c + d) <= 3) cout << 2 << endl;
        else if (abs(a - c) + abs(b - d) <= 6) cout << 2 << endl;
        else cout << 3 << endl;
    } else cout << 2 << endl;
    return 0;
}

D increment of coins

Description

题目链接

Solution

解法一：动态规划逆推
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F Valid payments

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