UVA - 138 Street Numbers
2013-10-12 19:08
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题意:有两个数分别是m,n,求前10组1 + 2 + ... + m-1 = m+1 + ... + n的m和n,用求和公式退出2*pow(m,2) = (n+1)*n,然后就是枚举成立的情况,当然因为数据少,也是可以打表节省时间
#include <iostream> #include <cmath> #include <cstring> #include <iomanip> #include <cstdio> using namespace std; const int MAXN = 1000000000; int main(){ long long a; double b; int count = 0; for (int i = 6; i < MAXN; i++){ b = (double) i * (i+1); b /= 2; b = sqrt(b); a = b; if (a == b){ cout << setw(10) << a << setw(10) << i << endl; ++count; } if (count == 10) break; } return 0; }
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