Codeforces 834 C The Meaningless Game(二分)
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题意:告诉你两个人的初始分都是1,他会进行若干盘游戏,每盘游戏选定这个自然数k,然后比赛,赢的人得到k^2分,输的人得到k分,然后告诉你终止分为a,b,问这个终止分是不是对的。
思路:因为两人都会是全部k的乘积的倍数,所以二分枚举全部k的乘积得到sum,如果sum^3=a*b,则那么这个值是对的,最后判断下是不是a和b都是这个的倍数以及sum^3=a*b就好了。
PS:r的范围为1e6的原因是,a和b的范围为1e9,所以a*b的范围为1e18,所以枚举1e6的范围就好了
#include <iostream>#include <cstring>#include <string>#include <queue>#include <vector>#include <map>#include <set>#include <stack>#include <cmath>#include <cstdio>#include <utility>#include <algorithm>#include <functional>#define N 100010#define LL long long#define inf 0x3f3f3f3f#define lson l,mid,ans<<1#define rson mid+1,r,ans<<1|1using namespace std;const LL mod = 1e9 + 7;const double eps = 1e-9;int main() { LL a, b, c; int n; while (~scanf("%d",&n)) { while (n--) { scanf("%lld %lld", &a, &b); c = a*b; LL l = 1, r = 1000005, mid; while (l <= r) { mid = (l + r) / 2; if (mid*mid*mid >= c) { r = mid - 1; } else { l = mid + 1; } } LL ans = r + 1; if (ans*ans*ans != c || (a%ans != 0) || (b%ans != 0)) { printf("No\n"); } else { printf("Yes\n"); } } } return 0;}
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