【google code jam Qualification Round 2017】【Oversized Pancake Flipper】【贪心】

题目大意

给你一列01串，每次可改变连续k个的状态，求全部变成1的最小代价。

解题思路

从左到右扫，要变就变，判断最后可不可行。

code

#include<set>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define ULL unsigned long long
#define min(a,b) ((a<b)?a:b)
#define max(a,b) ((a>b)?a:b)
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=Begin[j];i;i=Next[i])
using namespace std;
int const ml=1000+9,Inf=1e9;
int t,n,a[ml];
int main(){
freopen("d.in","r",stdin);
freopen("d.out","w",stdout);
scanf("%d",&t);
fo(cas,1,t){
char ch=getchar();a[0]=0;
while((ch!='-')&&(ch!='+'))ch=getchar();
while((ch=='-')||(ch=='+'))a[++a[0]]=(ch=='+'),ch=getchar();
scanf("%d",&n);int ans=0;
fo(i,1,a[0]-n+1)if(!a[i]){
ans++;
fo(j,i,i+n-1)a[j]=1-a[j];
}
int ok=1;
fo(i,1,a[0])if(!a[i]){ok=0;break;}
if(ok)printf("Case #%d: %d\n",cas,ans);
else printf("Case #%d: IMPOSSIBLE\n",cas);
}
return 0;
}