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
#

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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;
}