51nod 1087 1 10 100 1000
2016-09-19 21:58
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51nod 1087 1 10 100 1000
简单思维题,容易发现是1的下标为:1 2 4 7 11,它们的相邻差不断++,推出下标x符合:x = (n-1)*n/2+1;
简单思维题,容易发现是1的下标为:1 2 4 7 11,它们的相邻差不断++,推出下标x符合:x = (n-1)*n/2+1;
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <map> using namespace std; #define LL long long #define INF 0x3f3f3f3f #define PI acos(-1.0) #define E 2.71828 #define MOD 1000000007 #define N 1010 int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); n = (n-1)*2; int x = (int)sqrt(n); if( x*(x+1)== n) printf("1\n"); else printf("0\n"); } return 0; }
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