[C++] [ITSA竞赛题目] 计算斜正方形个数 (Counting Tilt Squares)

题目出自ITSA竞赛，解答仅供参考

题目连结: Counting Tilt Squares

解答:

#include <stdio.h>#include <stdlib.h>long long find(int r, int c);int main(void){printf("r=%d, c=%d, ans=%d\n", 2, 2, find(2, 2));             //output n=2,    ans=0printf("r=%d, c=%d, ans=%d\n", 3, 3, find(3, 3));             //output n=3,    ans=1printf("r=%d, c=%d, ans=%d\n", 4, 4, find(4, 4));             //output n=4,    ans=6printf("r=%d, c=%d, ans=%d\n", 5, 5, find(5, 5));             //output n=5,    ans=20printf("r=%d, c=%d, ans=%d\n", 6, 6, find(6, 6));             //output n=6,    ans=50printf("r=%d, c=%d, ans=%d\n", 7, 7, find(7, 7));             //output n=7,    ans=105printf("r=%d, c=%d, ans=%d\n", 8, 8, find(8, 8));             //output n=8,    ans=196printf("r=%d, c=%d, ans=%d\n", 9, 9, find(9, 9));             //output n=9,    ans=336printf("r=%d, c=%d, ans=%d\n", 10, 10, find(10, 10));         //output n=10,   ans=540printf("r=%d, c=%d, ans=%d\n", 100, 100, find(100, 100));     //output n=100,  ans=8004150system("pause");return 0;}long long find(int r, int c){long long sum = 0;int n = r < c ? r : c;for (int x = 1; x < n; x++){for (int y = x + 1; y < n - x ; y++){sum += (r - x - y) * (c - x - y) * 2;}}for (int x = 2; x <= n; x += 2){sum += (r - x) * (c - x);}/*for (int x = 1; x < n; x++){sum += (x + r - n) * (x + c - n);}*/return sum;}

结语:
因为没有正确解答可以比对，可以确定的只有前三个解，所以如果有错误或是有更好的解法都欢迎跟我说。