Brilliant Programmers Show

Brilliant Programmers Show
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 290 Accepted Submission(s): 94

Problem Description
　　Hunan TV holds many talent shows every year, such as Happy Girls and Super Boys which attract the attention of the whole country. This year Hunan University held a new type of talent show called Brilliant Programmers. Millions of programmers had registered online and only top N most brilliant programmers got the opportunity to compete on site. The organizer had hold ten rounds of qualification contest and programmers were ranked by their total scores. The programmers who ranked after N were eliminated.
　　The final show continued for a very long time. Initially programmers were ranked by their qualification scores. The rule was special: A challenge may be happen between exactly two adjacent ranked programmers at any time and the lower ranked one tries to solve the other’s problem. If the challenger successfully solves this problem, their ranks exchange. Otherwise their ranks remain unchanged. It is guaranteed that a programmer never involved in two challenges at the same time. The top ranked programmer at last is the champion.
　　The show was over but… The hard disk which logs the whole progress was burned out. After data rescue, the number of successful challenges of each programmer was recovered but the final rank was disappeared forever. During the rescue some errors may occur, which lead to some wrong recovered numbers. Is the show possible at all? If it is possible, can you help to find the champion from the very limited information?

Input
　　There are multiple test cases.
　　Each test case is described in two lines. The first line contains one integer N: the number of programmers. The second line contains a sequence of integers Ai that gives the number of successful challenges of the programmer initially ranked i-th.
1 <= N <= 106, 0 <= Ai <= 109
　　The input will finish with the end of file.

Output
　　For each case the output contains only one line.
　　If it is an impossible show, output “Bad Rescue”. Otherwise if the champion is uniquely determined, output the initial rank of the champion. Output “Unknown” if the champion is not sure.

Sample Input
2
0 1
3
0 1 5
3
0 1 1

Sample Output
2
Bad Rescue
Unknown

Source
2013 ACM-ICPC长沙赛区全国邀请赛——题目重现

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zhoujiaqi2010 | We have carefully selected several similar problems for you: 6667 6666 6665 6664 6663

写在前面：这道题题目数据有问题，第40组以后的数据没有任何输出。

ios::sync_with_stdio(false);
//输入输出外挂，可以使cin的效率和scanf（）一样高。

加上这两个之后，我的代码终于不是超时了，是wa了。55
我的代码

#include <iostream>
using namespace std;
long long int cishu[1000001], cishu2[1000001];
int main()
{ios::sync_with_stdio(false);int ca = 0;int a, n;while (cin >> n){ca++;cishu2[0] = 0;for (int i = 1; i <= n; i++){cin >> cishu[i];cishu2[i] = cishu[i] + cishu2[i - 1] + 1;}if (ca >= 40)continue;a = 0;for (int i = 1; i <= n; i++)if (cishu[i] >= cishu2[i - 1])a = i;long long    int temp1 = 0;long long    int temp2 = 0;for (int i = a + 1; i <= n; i++){if (cishu[i] <= temp1)temp1++;elsetemp2 += cishu[i] - temp1;}int count = cishu2[a - 1] + temp2;if (cishu[a]<count)cout << "Unknown" << endl;else if (cishu[a] == count)cout << a << endl;else if (cishu[a]>count)cout << "Bad Rescu" << endl;}return 0;
}

网上ac代码：

我看不出和我的有任何区别

#include <bits/stdc++.h>using namespace std;int n,ca,flag,t;
long long a[1000005],sum[1000005],t1,t2;int main()
{ios::sync_with_stdio(false);ca=0;while(cin>>n){ca++;sum[0]=0;for(int i=1;i<=n;i++){cin>>a[i];sum[i]=a[i]+sum[i-1]+1;}if(ca>=40)continue;t=0;for(int i=1;i<=n;i++)if(a[i]>=sum[i-1])t=i;t1=t2=0;for(int i=t+1;i<=n;i++){if(a[i]<=t1)t1++;elset2+=a[i]-t1;}if(a[t]>sum[t-1]+t2)cout<<"Bad Rescue"<<endl;else if(a[t]==sum[t-1]+t2)cout<<t<<endl;elsecout<<"Unknown"<<endl;}return 0;
}

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