题目描述

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include [HTML_REMOVED]

using namespace std;
int main()
{
char t;
cin >> t;
double q[12][12];
for (int i = 0; i < 12; i ){
for (int j = 0; j < 12; j ){
cin >> q[i][j];
}
}
double s = 0, c = 0;
for (int i = 1; i <= 11; i ){
for (int j = 12 - i; j <= 11; j ){
s += q[i][j];
c += 1;
}
}
if (t == ‘S’) printf(“%.1lf\n”, s);
else printf(“%.1lf\n”, s / c);
return 0;
}

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算法1

(暴力枚举) $O(n^2)$

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时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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