Hacked Exam 14pts 2021Google codejam round1A

https://codingcompetitions.withgoogle.com/codejam/round/000000000043585d/0000000000754750#analysis

只会前2个set，也就是n<=2

n=1时，比较一下s和q-s哪个大，如果q-s大就对答案全部取反期望得分就是q-s，否则就输出当前的答案，那么得分就是s

n=2a时，设得分分别为as,bs，a[i]!=b[i]的位置一共cnt个，那么这cnt个一定是一边得分一边不得分，那么可以计算出q-cnt这些相同的部分只得了sc1=(as+bs-cnt)/2分

所以相同的位置同样考虑取反得q-cnt-sc分还是不取反得sc，然后不同的位置考虑按a的来就是as-sc分，按b就是bs-sc分

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;const int maxl=510;int n,m,ans;
int as,bs;
char a[maxl],b[maxl],c[maxl];inline void prework()
{scanf("%d%d",&n,&m);scanf("%s",a+1);scanf("%d",&as);if(n==2){scanf("%s",b+1);scanf("%d",&bs);}
}inline void mainwork()
{if(n==1){int ares=m-as;ans=0;if(as>ares){for(int i=1;i<=m;i++)printf("%c",a[i]);printf(" ");printf("%d/1\n",as);}else{for(int i=1;i<=m;i++)printf("%c",'T'+'F'-a[i]);printf(" ");printf("%d/1\n",ares);}}else{int cnt=0;ans=0;for(int i=1;i<=m;i++)if(a[i]!=b[i])cnt++;if((as+bs-cnt)>m-cnt){for(int i=1;i<=m;i++)if(a[i]==b[i])c[i]=a[i];ans=(as+bs-cnt)/2;}else{for(int i=1;i<=m;i++)if(a[i]==b[i])c[i]='T'+'F'-a[i];ans=m-cnt-(as+bs-cnt)/2;}if(as>bs){ans+=as-(as+bs-cnt)/2;for(int i=1;i<=m;i++)if(a[i]!=b[i])c[i]=a[i];}else{ans+=bs-(as+bs-cnt)/2;for(int i=1;i<=m;i++)if(a[i]!=b[i])c[i]=b[i];}for(int i=1;i<=m;i++)printf("%c",c[i]);printf(" ");printf("%d/1\n",ans);}
}int main()
{int t;scanf("%d",&t);for(int i=1;i<=t;i++){prework();printf("Case #%d: ",i);mainwork();}return 0;
}

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Hacked Exam 14pts 2021Google codejam round1A

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