HDU4161 Iterated Difference

来源：互联网发布：linux打包解包编辑：程序博客网时间：2024/09/21 06:37

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4161

题意很简单，题也很水，训练时是队友做的，赛后自己写了写感觉没啥难度。

题意：给定一串数字，每次变化时每个数字等于这个数字与下一个数字的差的绝对值，问经过几次转变这一串数字变为相等的数字。直接模拟就好。

#include<iostream>#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<string>using namespace std;const int maxn = 25;int a[maxn];int main(){    int n,kase = 0;    while(scanf("%d",&n) == 1 && n)    {        memset(a, 0, sizeof(a));        int vis = 0;        for(int i = 0; i < n; i++)        {            scanf("%d",&a[i]);            if(i && a[i] == a[i-1])                vis++;        }        if(vis == n-1)        {            printf("Case %d: 0 iterations\n",++kase);            continue;        }        else        {            int a0,sum = 0,flag = 0;            a0 = a[0];            while(1)            {                if(sum == 1000) break;                for(int i = 0; i < n; i++)                {                    if(i == n-1) a[i] = abs(a[i] - a0);                    else a[i] = abs(a[i] - a[i+1]);                }                int ans = 0;                for(int j = 0; j < n-1; j++)                {                    if(a[j] == a[j+1])                        ans++;                }                flag = 0;                sum++;                if(ans == n-1)                {                    flag = 1;                    break;                }                else                {                    a0 = a[0];                }            }            if(flag)                printf("Case %d: %d iterations\n",++kase, sum);            else                printf("Case %d: not attained\n",++kase);        }    }        return 0;    }

0 0