UVaOJ 1225 – Digit Counting

问题描述

解法一：模拟（预处理）

可以提前打表 cnt 存储好所有的情况，一维索引表示输入数字 n ，二维索引范围为 0～9 的数位。

#include <bits/stdc++.h>

using namespace std;

int main() {
  const int MAX_VALUE = 10000, DIGIT_RANGE = 10;
  array<array<int, DIGIT_RANGE>, MAX_VALUE> cnt = {0};

  for (int i = 1; i < MAX_VALUE; ++i) {
    if (i > 1) cnt[i] = cnt[i - 1];
    int cur = i, digit;
    while (cur) {
      digit = cur % 10;
      cnt[i][digit]++;
      cur /= 10;
    }
  }

  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < DIGIT_RANGE; ++i) cout << (i ? " " : "") << cnt[n][i];
    cout << endl;
  }
  return 0;
}

#include <bits/stdc++.h>

using namespace std;

int main() {

const int MAX_VALUE = 10000, DIGIT_RANGE = 10;

array<array<int, DIGIT_RANGE>, MAX_VALUE> cnt = {0};

for (int i = 1; i < MAX_VALUE; ++i) {

if (i > 1) cnt[i] = cnt[i - 1];

int cur = i, digit;

while (cur) {

digit = cur % 10;

cnt[i][digit]++;

cur /= 10;

}

int t, n;

cin >> t;

while (t--) {

cin >> n;

for (int i = 0; i < DIGIT_RANGE; ++i) cout << (i ? " " : "") << cnt[n][i];

cout << endl;

}

return 0;

}

MAX_VALUE = 10000
DIGIT_RANGE = 10
cnt = [[0] * DIGIT_RANGE] * 2

for i in range(1, MAX_VALUE):
    if i > 1:
        cnt.append(cnt[i - 1].copy())
    cur = i
    while cur:
        digit = cur % 10
        cnt[i][digit] += 1
        cur //= 10

t = int(input())
for _ in range(t):
    n = int(input())
    for i in range(DIGIT_RANGE):
        print(cnt[n][i], end="\n" if i == 9 else " ")

MAX_VALUE = 10000

DIGIT_RANGE = 10

cnt = [[0] * DIGIT_RANGE] * 2

for i in range(1, MAX_VALUE):

if i > 1:

cnt.append(cnt[i - 1].copy())

cur = i

while cur:

digit = cur % 10

cnt[i][digit] += 1

cur //= 10

t = int(input())

for _ in range(t):

n = int(input())

for i in range(DIGIT_RANGE):

print(cnt[n][i], end="\n" if i == 9 else " ")

AOAPC II UVaOJ 字符串

UVaOJ 1225 - Digit Counting

问题描述

解法一：模拟（预处理）