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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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/**
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母题:P1548
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一、算正方形的个数
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枚举每一个格子,看以它为左上角的矩形共有多少个(正方形与长方形同属于矩形)
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二、算长方形个数(矩形=长方形+正方形)
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1.其实算长方形并不常见,但算矩形大家应该经常遇到,所以如果你会算矩形,再联系第一个问题,那答案就转化为 矩形个数-正方形个数.
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2.像求解正方形个数一样,固定矩形右下角(i,j),显然此时矩形个数为i*j.
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3.同理,求和即可.然后,再减去正方形的个数就是长方形的个数啦。
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*/
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LL n, m, s1, s2;
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int main() {
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cin >> n >> m;
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++) {
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s1 += min(i, j);//也可以理解为左上角
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s2 += i * j; //也可以理解为左上角开始,也是一样的
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}
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cout << s1 << " " << s2 - s1 << endl;
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return 0;
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}
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