Codeforces Round #433 Div. 2 B Maxim Buys an Apartment
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题意:
有n个屋子,其中k个屋子是有人住的,如果第i个屋子有人住,那么第i-1个屋子和第i+1个屋子就是特殊的,方便起见这里我们将特殊屋子的数量记为ans。然后给你n和k,让你输出最小的ans和最大的ans。
失误 总结:考虑不全面,没有逻辑关系
失误点:每个已经居住的房子要想周围是空的,达到最佳状态,则需要足够的地方,即3*k <= n,此时得到最大值2*k,否则得到 n-k,解题时要有足够的逻辑分析支撑
失误点2:此题数据范围1e9,看似不超int 2e10,但是实际运算过程中k*3则会超过int,所以要特别注意诸如此类的数据类型计算会爆int的情况,本场比赛c题也是这个问题
AC code:
#include<cstdio>#include<algorithm>#include<cstring>#include<vector>#include<queue>#include<cmath>#include<iostream>using namespace std;int main(){ long long n,k; cin>>n>>k; if(k==0) { printf("%d %d\n",k,k); } else if(k==n) { printf("0 0\n"); } else { int u; if(k*3<=n) u = k*2; else u = n-k; printf("1 %d\n",u); } return 0;}
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