cdoj第13th校赛初赛F - Fabricate equation

http://acm.uestc.edu.cn/#/contest/show/54

 

F - Fabricate equation

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

Given an integer Y, you need to find the minimal integer K so that there exists a X satisfying X−Y=Z(Z≥0) and the number of different digit between X and Z is K under decimal system.

For example: Y=1, you can find a X=100 so that Z=99 and K is 3 due to 1≠0 and 0≠9. But for minimization, we should let X=1 so that Z=0 and K can just be 1.

Input

Only one integer Y(0≤Y≤1018).

Output

The minimal K.

Sample input and output

Sample Input	Sample Output
1	1
191	2

 

 

思路：贪心。倒着扫描，遇到0就忽略，因为对应被减数的该位也设0就好；遇到9，这个特殊，因为比如290-191=99，后面进位后，9这个位也可以使得被减数与结果的该位相同，这样的情况需要两个条件：必须后面可以进位。假如减数那位为0，不论结果为什么，都无法产生进位。第二个条件是，被减数的前一位可以借位，也就是说9这种情况完成后，即便前面遇到减数那位为0，也不能再忽略，只能当一般情况处理。一般情况自然就是k++。

官方题解：

代码：

#include <fstream>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <vector>

using namespace std;

#define PI acos(-1.0)
#define EPS 1e-10
#define lll __int64
#define ll long long
#define INF 0x7fffffff

char c[25];

int main(){
    //freopen("D:\\input.in","r",stdin);
    //freopen("D:\\output.out","w",stdout);
    scanf("%s",c+1);
    int l=strlen(c+1),k=0;
    c[l+1]='0';c[0]='0';
    for(int i=l;i>0;i--){
        if(c[i]=='0')   continue;
        if(c[i]=='9'){
            if(c[i+1]=='0'){
                k++;
                continue;
            }else{
                if(c[i-1]=='0') c[i-1]++;//把前面的0毁掉
                continue;
            }
        }else{
            k++;
        }
    }
    if(c[0]!='0')   k++;
    printf("%d\n",k);
    return 0;
}