首页 > ACM题库 > HDU-杭电 > HDU 4134-Sequence Folding-模拟-[解题报告]HOJ
2015
04-16

HDU 4134-Sequence Folding-模拟-[解题报告]HOJ

Sequence Folding

问题描述 :

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.

StrangeStandard

Figure 1.a Before Folding

StrangeStandard

Figure 1.b After one step of folding

StrangeStandard

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.

输入:

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.

输出:

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.

样例输入:

2
5
2 5 10 3 -4
3
5 4 -3

样例输出:

Case #1: Alice
Case #2: Bob

模拟不解释

int a[110];
int main(){
    int t;
    scanf("%d",&t);
    int ca = 1;
    while(t--){
        int n;
        scanf("%d",&n);
        int i,j;
        for(i=1;i<=n;i++)scanf("%d",&a[i]);
        int win = 0;
        i = 1,j = n;
        while(1){
            if(j-i+1==2){
                if(a[1]>a[2])win = 1;
                else win = 2;
                break;
            }
            int ii = i,jj = j;
            while(ii<=jj){
                a[ii++] += a[jj--];
            }
            i = 1,j = (j+1)/2;
        }
        printf("Case #%d: ",ca++);
        if(win==1)puts("Alice");
        else if(win==2)puts("Bob");
    }
    return 0;
}

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参考:http://blog.csdn.net/leolin_/article/details/7306748


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