【codeforces】Vacations
2016-07-24 00:04
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本题要考虑清楚各种情况,其实也很简单一个循环解决了,重要的还是贪心的思想。尽可能多的不休息。
***1.遇到0,休息天数加1;
2.遇到非第0(第0个元素是三可以舍掉,不影响结果)个元素是3的,如
果前面是2,就让它等1,反之等2;
3.遇到两个1或2相等,后一个等0,休息天数加1。*
#include<stdio.h> int a[102]; int main() { int n; while(scanf("%d",&n)!=EOF) { int sum=0; for(int l=0; l<n; l++) { scanf("%d",&a[l]); if(a[l]==0) { sum++; } else if(a[l]==3&&l) { if(a[l-1]==1) a[l]=2; else if(a[l-1]==2) a[l]=1; } else if(a[l]==a[l-1]) { a[l]=0; sum++; } } printf("%d\n",sum); } return 0; }
http://acm.hust.edu.cn/vjudge/contest/123357#problem/I
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