String Builder

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You are going to read four numbers: n, a, b and c, like this:

12 2 5 3

First, n is used to build up a string from 0 to n, like this:

0123456789101112

is a string build up for n=12.

Then, in all the digits from index a to index b, count the appearence of c.

For the string above, 2 5 is:

2345

Thus the appearence of 3 is 1.

Input Format:

Four positive numbers, n, a, b and c, where a<b<n<10000, and 0<=c<=9..

Output Format:

One number represnets the length of the generated string. One number represents the apprence of c. There is a space between the two numbers.

Sample Input:

12 2 5 3

结尾无空行

Sample Output:

16 1

结尾无空行

 

代码如下

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#include<stdio.h>
#include<vector>
using namespace std;
int main()
{
    int n,a,b,c;
    scanf("%d %d %d %d",&n,&a,&b,&c);
    int j=0,l=0,sum=0;
    for(int i=0;i<=n;i++)
    {
        if(!i)//特判0是不是满足条件
        {
            if(a==0&&c==0)
                sum++;
            l++;
            continue;
        }
        j=i;
        vector<int> v;
        while(j)//将数分成1个个0-9的数
        {
            v.push_back(j%10);
            j/=10;
        }
        for(int j=v.size()-1; j>=0; j--)//记录c在a-b位置出现的次数
        {
            if(l>=a&&l<=b&&v[j]==c)
                sum++;
            l++;
        }
    }
    printf("%d %d",l,sum);//输出长度和总数
    return 0;
}

emmm，比赛时，吃了英语的亏

这篇文章以互相学习为主，有什么错的还望告知，谢谢啦