B-number

Description
A wqb-number, or B-number for short, is a non-negative integer whose decimal form contains the sub- string "13" and can be divided by 13. For example, 130 and 2613 are wqb-numbers, but 143 and 2639 are not. Your task is to calculate how many wqb-numbers from 1 to n for a given integer n.

Input
Process till EOF. In each line, there is one positive integer n(1 <= n <= 1000000000).

Output
Print each answer in a single line.

Sample Input
13
100
200
1000

Sample Output
1
1
2

2

题意：与“不要62”相反，即“要13”

#include<stdio.h>
#include<string.h>

bool flag;
int now, len;
int numb[30];                     //存储数字
int note[30][20][20];

int dfs(int now, bool flag, int left, int key){
    if(now == len){
        if(!left && key == 2) return 1;        //能被13整除，有13
        return 0;
    }
    if(flag == false && note[now][left][key] != -1) return note[now][left][key];
    int bound = 9;
    if(flag) bound = numb[now + 1];//bound = flag ? num[now + 1] : 9;         //如果当前位数到边界下一位数范围存在约束（0～numb[now + 1])否则范围无约束（即0~9)
    int ret = 0;
    for(int i = 0; i <= bound; i++){
        int lleft = left*10 + i;
        int k = 0;
        lleft %= 13;
        if(i == 1) k = 1;
        if(key == 1 && i == 3) k = 2;            //有13
        if(key == 2) k = 2;
        ret += dfs(now + 1, flag == true && i == bound, lleft, k);
    }
    if(flag == false) note[now][left][key] = ret;
    return ret;
}

int main(){
    int na, n;
    char a[30];
    while(~scanf("%d", &na)){
        sprintf(a, "%d", na);
        n = 0;
        len = strlen(a);
        for(int i = 0; i < len; i++) numb[i + 1] = a[i] - '0';
        memset(note, -1, sizeof(note));
        n = dfs(0, true, 0, 0);                //计算0~na之间的13的数的个数
        printf("%d\n", n);                        //sum即1~na之间有13的数的个数
    }
    return 0;
}