程序代码：

算法实现题1-1
1.问题描述:一本书的页码从自然数1开始顺序编码直到自然数n.书的页码按照通常的
        习惯编排,每个页码都不含多余的前导数字0.例如,第6页用数字6表示,而不是06或者006
        等.数字计数问题要求对给定书的总页码n,计算出书的全部页码中分别用到多少次数字
        0，1，2，....，9。
2.算法设计:给定表示书的总页码的十进制整数n(1<=n<=10^9),计算书的全部页码中分别
            用到多少次数字0,1,2.....9.
3.数据输入:输入数据由文件名input.txt的文本文件提供.每个文件只有一行,给出表示书
            的总页码的整数n
4.结果输出:将计算结果输出到文件output.txt.输出文件共10行,在第k行输出页码中用到数字k-1
            的次数,k=1,2,....10.
            输入文件示例                     输出文件示例
            input.txt                           output.txt
            11                                  1
                                                4
                                                1
                                                1
                                                1
                                                1
                                                1
                                                1
                                                1
                                                1 

程序代码：

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

int main() {
    int n, a[10], i, j, b, c;
    while(scanf("%d", &n) == 1) {
        memset(a, 0, sizeof(a));
        for(i = 1; i <= n; i++) {
            c = i;
            while(c) {
                b = c % 10; c /= 10;
                for(j = 0; j <= 9; j++)
                    if(b == j) {a[j]++; break;}
            }
        }
        for(i = 0; i <= 9; printf("%d\n", a[i++])) ;
    }
    return 0;
}

程序代码：

#include<stdio.h>

int f(int n, int x)
{
        int count = 0;
        int factor = 1;
        int lower_num = 0;
        int cur_num = 0;
        int higher_num = 0;

        while ((x!=0&&n/factor!=0) || (x==0&&n/factor>=10))
        {
                lower_num = n - n/factor*factor;
                cur_num = n/factor%10;
                higher_num = n/factor/10;

                if (x==0)
                        higher_num--;
                if (cur_num < x)
                        count += higher_num*factor;
                if (cur_num == x)
                        count += higher_num*factor + lower_num + 1;
                if (cur_num > x)
                        count += (higher_num+1) * factor;
                
                factor *= 10;
        }
        return count;
}

int main(void)
{
        int n;
        int i;
        int result[10];
        for (i=0; i<10; i++)
                result[i] = 0;
        scanf("%d", &n);
        for (i=0; i<10; i++)
        {
                result[i] = f(n, i);
                printf("%d\n", result[i]);
        }

        return 0;
}

程序代码：

#include <stdio.h>
#include <string.h>
int main() {
    int n, len, a[10], b[100], c[100], t, i, j, d[100], f, e;
    while(scanf("%d", &n) == 1) {
        len = 0; t = n;
        if(n == 0) a[0] = 0;
        memset(a, 0, sizeof(a));
        d[0] = 1;
        while(t) {len++; t /= 10;}
        for(i = 1; i <= len; i++) {
            d[i] = 1;
            for(j = 1; j <= i; j++) d[i] *= 10;
            b[i] = n / d[i];
            c[i] = n % d[i];
        }
        for(i = 1; i <= len; i++)
            for(j = 1; j <= 9; j++)
                a[j] += b[i] * d[i - 1] + (c[i] / d[i - 1] > j ? d[i - 1] : (c[i] / d[i - 1] == j ? (c[i] % d[i - 1] + 1) : 0));
        for(i = 1; i <= n; i++) {
            f = i;
            while(f) {
                e = f % 10; f /= 10;
                if(!e) a[0]++;
            }
        }
        for(i = 0; i <= 9; printf("%d\n", a[i++])) ;
    }
    return 0;
}