Google Code jam Qualification Round 2015 --- Problem A. Standing Ovation

Problem A. Standing Ovation 

Problem's Link:   https://code.google.com/codejam/contest/6224486/dashboard#s=p0

---

 

Mean: 

题目说的是有许多观众，每个观众有一定的羞涩值，只有现场站起来鼓掌的人数达到该值才会站起来鼓掌，问最少添加多少羞涩值任意的人，才能使所有人都站起来鼓掌。

 

analyse:

贪心模拟一下，从前往后扫一遍就行。

Time complexity: O(n)

 

Source code: 

 

#include<iostream>

#include<cstdio>

#include<cmath>

using namespace std;



const int MAXN=1110;

int n;

char s[MAXN];

int main()

{

//        freopen("E:\\Code_Fantasy\\C\\A-small-attempt0.txt","r",stdin);

//        freopen("E:\\Code_Fantasy\\C\\A-small-attempt1.txt","w",stdout);

        int t;

        scanf("%d",&t);

        for(int Cas=1;Cas<=t;++Cas)

        {

                scanf("%d",&n);

                scanf("%s",s);

                int ans=0;

                int shy=s[0]-'0';

                for(int i=1;i<=n;++i)

                {

                        if(s[i]-'0'!=0)

                        {

                                if(shy>=i)

                                        shy+=(s[i]-'0');

                                else

                                {

                                        ans+=(i-shy);

                                        shy=i+(s[i]-'0');

                                }

                        }

                }

                printf("Case #%d: %d",Cas,ans);

                if(Cas!=t) puts("");

        }

        return 0;

}

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