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PAT 1049 Counting Ones

时间:2014-11-08 21:57:28      阅读:269      评论:0      收藏:0      [点我收藏+]
#include <cstdio>
#include <cstdlib>

using namespace std;

#define COL 10
#define ROW 10

int tbl[ROW][COL];

void print(int* tbl) {
    for (int i=0; i<ROW; i++) {
        for (int j=0; j<COL; j++) {
            printf("%11d", tbl[i * COL + j]);
        }
        printf("\n");
    }
}

int ones_aux(int num, int digits, int tens) {
    if (digits < 1 || tens < 1) return 0;
    int first = num / tens;
    int sum = 0;

    for (int i=0; i<first; i++) {
        sum += tbl[digits-1][i];
    }

    if (first == 1) {
        sum += num % tens + 1;
    }
    int subsum = ones_aux(num % tens, digits - 1, tens / 10);
    return sum + subsum;
}

int ones(int num) {
    int digits = 1;
    int tmp = num;
    int tens = 1;
    
    while (tmp) {
        tmp /= 10;
        if (tmp) {
            tens *= 10;
            digits++;
        }
    }

    return ones_aux(num, digits, tens);
}

int main() {
    int tens = 1;
    int sum = 0;
    for (int i=0; i<ROW; i++) {
        for (int j=0; j<COL; j++) {
            if (j == 0) {
                tbl[i][j] = sum;
                continue;
            }
            if (j == 1) {
                tbl[i][j] = tens * 1;
            }
            tbl[i][j] += tbl[i][0];
            sum += tbl[i][j];
        }
        tens *=10;
    }
    int n;
    scanf("%d", &n);
    printf("%d\n", ones(n));
    
    system("pause");
    return 0;
}
          0          1          0          0          0          0          0          0          0          0
          1         11          1          1          1          1          1          1          1          1
         20        120         20         20         20         20         20         20         20         20
        300       1300        300        300        300        300        300        300        300        300
       4000      14000       4000       4000       4000       4000       4000       4000       4000       4000
      50000     150000      50000      50000      50000      50000      50000      50000      50000      50000
     600000    1600000     600000     600000     600000     600000     600000     600000     600000     600000
    7000000   17000000    7000000    7000000    7000000    7000000    7000000    7000000    7000000    7000000
   80000000  180000000   80000000   80000000   80000000   80000000   80000000   80000000   80000000   80000000
  900000000 1900000000  900000000  900000000  900000000  900000000  900000000  900000000  900000000  900000000

先生成0~9, x0~x9, x00~x99, x000~x999...这这些范围内的1数量表,然后把输入的数组一步一步拆分到这些区域上如321 = 00~099 + 100~199 + 200~299 + 00~09 + 10~19 + 0~1,当然对于开头为1的数字需要额外加上一些

PAT 1049 Counting Ones

原文:http://www.cnblogs.com/lailailai/p/4083983.html

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