10082 (2/5)
| #include <stdio.h> char sen[1000]; char kb[] = "`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./"; int main(){ while(gets(sen)){ for(int i = 0; sen[i]; i++){ if(sen[i] == ' ') printf(" "); else{ int j = 0; while(sen[i] != kb[j+1]) j++; printf("%c", kb[j]); } } printf("\n"); } return 0; } |
10189 (2/5)
| #include <stdio.h> #include <stdlib.h> int main(){ int w, h; int f_n = 0; char f[101][101]; int count[101][101]; while(scanf("%d %d", &h, &w)){ if(h == 0 || w == 0 || h > 100 || w > 100) break; f_n++; for(int i = 0; i < h; i++){ for(int j = 0; j < w; j++){ f[i][j] = '0'; count[i][j] = 0; } } for(int i = 0; i < h; i++){ for(int j = 0; j < w; j++){ f[i][j] = getchar(); if(f[i][j] == '\n') f[i][j] = getchar(); } } printf("Field #%d:\n", f_n); for(int i = 0; i < h; i++){ for(int j = 0; j < w; j++){ if(f[i][j] == '*'){ if(i == 0 && j == 0){ count[i][j + 1]++; count[i + 1][j]++; count[i + 1][j + 1]++; } else if(i == 0 && j > 0 && j != w){ count[i][j + 1]++; count[i][j - 1]++; count[i + 1][j - 1]++; count[i + 1][j]++; count[i + 1][j + 1]++; } else if(i == 0 && j == w){ count[i][j - 1]++; count[i + 1][j]++; count[i + 1][j - 1]++; } else if(i > 0 && j == 0){ count[i][j + 1]++; count[i - 1][j]++; count[i - 1][j + 1]++; count[i + 1][j]++; count[i + 1][j + 1]++; } else if(i > 0 && j > 0 && j != w){ count[i][j - 1]++; count[i][j + 1]++; count[i - 1][j - 1]++; count[i - 1][j]++; count[i - 1][j + 1]++; count[i + 1][j - 1]++; count[i + 1][j]++; count[i + 1][j + 1]++; } else if(i > 0 && j == w){ count[i][j - 1]++; count[i - 1][j]++; count[i - 1][j - 1]++; count[i + 1][j]++; count[i + 1][j - 1]++; } else if(i == h && j == 0){ count[i][j + 1]++; count[i - 1][j]++; count[i - 1][j + 1]++; } else if(i == h && j > 0 && j != w){ count[i][j - 1]++; count[i][j + 1]++; count[i - 1][j - 1]++; count[i - 1][j]++; count[i - 1][j + 1]++; } else{ count[i][j - 1]++; count[i - 1][j]++; count[i - 1][j - 1]++; } } } } printf("\n"); for(int i = 0; i < h; i++){ for(int j = 0; j < w; j++){ printf("%c", f[i][j]); } printf("\n"); } printf("\n"); for(int i = 0; i < h; i++){ for(int j = 0; j < w; j++){ if(f[i][j] == '*'){ printf("*"); } else printf("%d", count[i][j]); } printf("\n"); } printf("\n"); } return 0; } |
10082需要知道的就是像這種題目就要先建立一個字元陣列去模擬題目的內容，如鍵盤。
10189我是用土法煉鋼，主要是我對陣列結構不熟悉所以就用比較穩的方式去做地雷數量判定。
然後非常感謝豆漿、啊雪、啊柴學長的協助不然這題真的做不出來……
題目要截圖很麻煩佔空間，想知道的就直接去OJ官網看相對應題號就好了。
達人
