題意: 給出一張地圖,每次只能向右或向下移動並得到該位置數字的值,試問從(x1,y1)到(x2,y2)的路徑和有多少種?
思路:
- 不難發現地圖的規律,從右上角到左下角數字依次增加1
- 當然也會發現,若先選定一條路徑得到和爲x,那麼要得到x+1的和便需要選擇路徑中一個點替換成其相應左下方那個點。
- 於是便能得到規律,和的種類數有|x1 - x2| * |y1 - y2| + 1
代碼實現:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int t, x1, x2, y1, y2;
signed main()
{
IOS;
cin >> t;
while(t --){
cin >> x1 >> y1 >> x2 >> y2;
cout << abs(x1 - x2) * abs(y1 - y2) + 1 << endl;
}
return 0;
}