2017 Multi-University Training Contest - Team 5 Rikka with Competition
2017-08-18 16:13
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http://acm.hdu.edu.cn/showproblem.php?pid=6095
题意:
给出n个数,进行n−1场比赛,每场比赛如果双方差值的绝对值大于k,那么数值较大的人赢,否则都有可能赢,现在要你求最后最多有多少人可能赢。思路:
先把值排个序,然后判差值是否大于k,第一次大于k之前的数的个数就是答案。代码:
#include<bits/stdc++.h> using namespace std; int a[1000005]; int main(){ int n,k; int T; while(~scanf("%d",&T)){ while(T--){ scanf("%d %d",&n,&k); for(int i = 0;i < n;i++){ scanf("%d",&a[i]); } int ans = n; sort(a,a + n); for(int i = n - 1;i >= 0;i--){ if(a[i] - a[i - 1] > k){ ans -= i; break; } } printf("%d\n",ans); } } }
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