nenu contest
2014-08-20 21:58
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http://vjudge.net/vjudge/contest/view.action?cid=54393#overview
A n^2能过 对第二个n我二分了一下,快了一点点,nlogn
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A n^2能过 对第二个n我二分了一下,快了一点点,nlogn
#include<cstdio> typedef __int64 LL; const int M=1000010; char a[M][4]; int main(){ int n; while(~scanf("%d",&n)){ for(int i=0;i<n;i++){ scanf("%s",a[i]); } LL ans=0; LL sum=0; for(int i=n-1;i>=0;i--){ if(a[i][0]=='W') ans+=sum; else sum++; } printf("%I64d\n",ans); } return 0; }
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