codeforces 716 B. Anatoly and Cockroaches(greedy)
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题意:给一些在一排的蟑螂涂色或者两两交换位置,使最终的颜色交替出现,输出最小步数
枚举两种r和b在开头位置的情况,取最小,推一下计算公式就出来了
#include<cstring>#include<string>#include<iostream>#include<queue>#include<cstdio>#include<algorithm>#include<map>#include<cstdlib>#include<cmath>#include<vector>//#pragma comment(linker, "/STACK:1024000000,1024000000");using namespace std;#define INF 0x3f3f3f3fchar col[100006];int n;int solve(char c){ int sam=0,dif=0; for(int i=0;i<n;i+=2) { if(col[i]==c) sam++; } for(int i=1;i<n;i+=2) { if(col[i]==c) dif++; } int ans=min((((n+1)/2)-sam),dif); return dif+(((n+1)/2)-sam)-ans;}int main(){ scanf("%d",&n); scanf("%s",col); int ans1=solve('r'); int ans2=solve('b'); printf("%d\n",min(ans1,ans2)); return 0;}
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