HDU 4272 LianLianKan（模拟）

LianLianKan

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3292    Accepted Submission(s): 996

Problem Description

I like playing game with my friend, although sometimes looks pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.
Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with one same-value element. After pop them from stack, all left elements will fall down. Although the game seems to be interesting, it's really naive indeed. 

To prove I am a wisdom among my friend, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 6 with it. 
Before the game, I want to check whether I have a solution to pop all elements in the stack.

 

Input

There are multiple test cases.
The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)
The next line contains N integer ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)

 

Output

For each test case, output “1” if I can pop all elements; otherwise output “0”.

 

Sample Input

21 131 1 121000000 1

 

Sample Output

100

 

Source

2012 ACM/ICPC Asia Regional Changchun Online

    题意:给出一个序列，其中距离不超过6的两个相同的数字可以消除掉，现在问把能消除的全部消除之后序列中是不是还有数字，有的话输出0，没有的话输出1

思路：按照题意模拟一遍然后判断是不是为空就可以了

点击打开链接

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#include<queue>#include<stack>#include<vector>#include<map>using namespace std;int n;__int64 a[10010];int v[10010];__int64 b[10010];int main() {    while(scanf("%d",&n)!=EOF) {        for(int i=1; i<=n; i++) {            scanf("%I64d",&a[i]);        }        memset(v,0,sizeof(v));        int t = 0;        int flag;        for(int i=1; i<=n; i++) {            int pt = t;            flag = 0;            if(v[i] == 1){continue;            }            for(int k=1; k<=t; k++) {                if(a[i] == b[k]) {                    for(int pi=k; pi<t; pi++) {                        b[pi] = b[pi+1];                    }                    t--;                    flag = 1;                }            }            if(pt == t) {                int pf = 0;                for(int j=i+1; j<=i+5 && j<=n; j++) {                    if(a[i] == a[j]) {                        v[i] = 1;                        v[j] = 1;                        pf = 1;                        break;                    }                }                if(pf == 0) {                    b[++t] = a[i];                }            }        }        if(t == 0) {            printf("1\n");        } else {            printf("0\n");        }    }    return 0;}