Gym 100712I Bahosain and Digits(枚举)

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Gym 100712I


Bahosain and Digits


Bahosain has a string of digits. He is going to perform the following operation on the string as many times as
needed to make all digits in the string the same.
In one move, he can choose a substring of length Kand change each digit in this substring to it’s next digit,
the digit that goes after ‘9’is ‘0’.
For example, applying the operation on the string 905 will change it to 016.
Your task is to determine the maximum K?such that it is possible to make all digits in the string equal using
the described operation.




Input
The first line of input contains an integer N (1 ≤ T ≤ 128)that represents the number of test cases.
Each test case contains a string of no more than 250 digits (0 - 9).
Each string contains at least two different digits.




Output
For each test case print a single line with the maximum possible K.




Sample Input
3
04
651
0552




Sample Output
1
2
3




Note
In the last test case, when K= 3,Bahosain can perform the operation on the substring 552 five times to get
the string 0007,then perform the operation again on the substring 000seven times to get 7777.




题目大意就是给一个序列，每次可以转动一个区间，使这个区间的值加1,9则变成0，目标是是这个序列的值最终一样，求每次转动的区间的最大长度。
可以先从大到小枚举区间长度ans，同时枚举最终整个序列的值k，在循环体内计算每一位要转动的次数，同时对后(ans-1)位进行特殊判断，如果当前情况合法，则后(ans-1)位需要转动的次数都为0。循环体内不能每次修改ans个值，会造成超时，所以循环体内的时间复杂度要是O(n)。因此我们可以定义一个数组change，change[i]表示经过前i-1次转动第i个位置要转动为k所要转动的次数，再另外一个变量sum表示对第i个位置造成影响的转动次数，即从i-ans位到i-1位转动次数之和。


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#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long LL;const int MAXN=260;char str[MAXN];int num[MAXN],n;int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif // ONLINE_JUDGE    int tcase;    scanf("%d",&tcase);    while(tcase--)    {        scanf("%s",str);        n=strlen(str);        for(int i=0;i<n;i++)        {            num[i]=str[i]-'0';        }        int ans;        for(ans=n;ans>=1;ans--)        {            int flag;            for(int k=0;k<10;k++)            {                int sum=0,change[MAXN]={0};                for(int i=0;i<n-(ans-1);i++)//循环到n-(ans-1)，最后(ans-1)位要进行特殊判断                {                    if(i-ans>=0)//防止下标越界                    {                        sum-=change[i-ans];                    }                    change[i]=((k+10)-(num[i]+sum)%10)%10;//k+10为了防止k比num[i]小                    sum+=change[i];                }                flag=1;                for(int i=n-(ans-1);i<n;i++)                {                    if(i-ans>=0)                    {                        sum-=change[i-ans];                    }                    change[i]=((k+10)-(num[i]+sum)%10)%10;                    if(change[i]!=0)//如果当前情况合法，则后(ans-1)位都不需要转动                    {                        flag=0;                        break;                    }                }                if(flag)                {                    printf("%d\n",ans);                    break;                }            }            if(flag)                break;        }    }    return 0;}