Codeforces Round #311(Div.2) (2/5)
A Ilya and Diplomas
Soon a school Olympiad in Informatics will be held in Berland, n schoolchildren will participate there.
At a meeting of the jury of the Olympiad it was decided that each of the n participants, depending on the results, will ge >>继续阅读
Codeforces Round #310(Div.2) (3/5) (Div.1) (1/5)
A.Case of the Zeros and Ones
Andrewid the Android is a galaxy-famous detective. In his free time he likes to think about strings containing zeros and ones.
Once he thought about a string of length n consisting of zeroes and ones. Consider the fol >>继续阅读
Codeforces Round #309(Div.2) (3/5) (Div.1) (1/5)
A.Kyoya and Photobooks
Kyoya Ootori is selling photobooks of the Ouran High School Host Club. He has 26 photos, labeled "a" to "z", and he has compiled them into a photo booklet with some photos in some order (possibly with some photos being duplic >>继续阅读
BestCoder Round #45 (2/4)
A.Dylans loves numbers
Problem Description
Who is Dylans?You can find his ID in UOJ and Codeforces. His another ID is s1451900 in BestCoder. And now today's problems are all about him.Dylans is given a number N. >>继续阅读
Codeforces Round #308(Div.2) (5/5)
A.Vanya and Table
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.
In this table, Vanya chose n re >>继续阅读
2015年四川省ACM省赛(1/10)
A题
由于正的无理数和任何正数相加都是无理数,所以只要判断所给的数是否都是完全平方数即可
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#include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> #define lson rt << 1, l, m #define rson rt << 1 | 1, m + 1, r using namespace std; const int INF = 0x7FFFFFFF; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int MOD = 7; const int maxn = 100000 + 66; int a[maxn]; int main() { // ifstream cin ("aaa.txt"); // freopen ("aaa.txt", "r", stdin); int n; while (scanf ("%d", &n) != EOF) { for (int i =1;i <= n ; i++) scanf ("%d", &a[i]); bool is_valid = true; for (int i = 1; i <= n; i++) { int b = (int) sqrt ((double) a[i]); if (b * b != a[i]) { is_valid = false; break; } } if (is_valid) puts ("Yes"); else puts ("No"); } return 0; } |