我的PAT-BASIC代碼倉:https://github.com/617076674/PAT-BASIC
原題連結:https://pintia.cn/problem-sets/994805260223102976/problems/994805266514558976
題目描述:
![](https://img.laitimes.com/img/__Qf2AjLwojIjJCLyojI0JCLiAzNfRHLGZkRGZkRfJ3bs92YscjMfVmepNHLykEVPhXTq1EeRpHW4Z0MMBjVtJWd0ckW65UbM5WOHJWa5kHT20ESjBjUIF2X0hXZ0xCMx81dvRWYoNHLrdEZwZ1Rh5WNXp1bwNjW1ZUba9VZwlHdssmch1mclRXY39CXldWYtlWPzNXZj9mcw1ycz9WL49zZuBnL0YTN2QDNzkDM3IDMxgTMwIzLc52YucWbp5GZzNmLn9Gbi1yZtl2Lc9CX6MHc0RHaiojIsJye.png)
知識點:格式化輸出
思路:按題述程式設計即可
輸出資料的時候需要格式化輸出,前面補0直至數字長度達到3位。
時間複雜度是O(M * N)。空間複雜度是O(1)。
C++代碼:
#include<iostream>
using namespace std;
int main(){
int M, N, A, B, newColor;
scanf("%d %d %d %d %d", &M, &N, &A, &B, &newColor);
int tempColor;
for(int i = 0; i < M; i++){
for(int j = 0; j < N; j++){
scanf("%d", &tempColor);
if(tempColor >= A && tempColor <= B){
tempColor = newColor;
}
printf("%03d", tempColor);
if(j != N - 1){
printf(" ");
}
}
printf("\n");
}
}
C++解題報告: