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

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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;}

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