HDU4134:Sequence Folding

Problem Description

Alice and Bob are practicing hard for the new ICPC season. They hold many private contests where only the two of them compete against each other. They almost have identical knowledge and skills, the matter which results many times in ties in both the number of problems solved and in time penalty! To break the tie, Alice and Bob invented a tie breaker technique called sequence folding! The following are the steps of this technique:
1- Generate a random integer N >= 2.
2- Generate a sequence of N random integers.
3- If N = 2 go to step 6.
4- Fold the sequence by adding the Nth element to the first, the N-1th element to the second and so on, if N is odd then the middle element is added to itself, figure 1 illustrates the folding process.
5- Set N = ceil (N/2) and go to step 3.
6- The sequence now contains two numbers, if the first is greater than the second then Alice wins, otherwise Bob wins.


Figure 1.a Before Folding

Figure 1.b After one step of folding

Figure 1.c After two steps of folding, Alice wins!

In this problem you’re given the sequence of N integers and are asked determine the contest winner using the sequence folding tie breaker technique.

 

Input

The first line contains T (1 <= T <= 100), the number of test cases. The first line of each test case contains an integer (2 <= N <= 100), the number of elements of the sequence. The next line contains N space separated integers. The sum of any subset of the numbers fit in a 32 bit signed integer.

 

Output

For each test case print the name of the winner. Follow the output format below.

 

Sample Input

252 5 10 3 -435 4 -3

 

Sample Output

Case #1: AliceCase #2: Bob

 

水题一道，按照题目的意思来就可以了

意思就是把一个序列对折，对应的数字相加

如果是奇数，那么对折后中间的数就自己加自己

最后只剩下两个

如果第一个比第二个打，那么就是Alice胜

 

#include <stdio.h>int main(){    int t,cas = 1;    int a[105],i,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i = 0;i<n;i++)        scanf("%d",&a[i]);        while(n!=2)        {            for(i = 0;i<n;i++)            a[i] = a[i]+a[n-i-1];            if(n%2)            n = (n+1)/2;            else            n/=2;        }        printf("Case #%d: ",cas++);        if(a[0]>a[1])        printf("Alice\n");        else        printf("Bob\n");    }    return 0;}