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Every year,prince prepares a birthday gift for princess.The gift is a box,which is decorated with jewel.
As princess grows older,the level of the jewelry box gradually increase from A to Y.
Every level N jewelry box will be packed in level (N-1) jewelry box(except level 1 jewelry box).
For example,when princess was 1 year old,she got a level-A (level 1) jewelry box.When pricess was 2 years old,she got a level-B(level 2) jewelry box.The level-B(level 2)jewelry box was packed in level-A(level 1) jewelry box.
Because prince has obsessive-compulsive disorder, there will be a paper box between two jewelry box.
Please help prince to determine the birthday gift.
The first line contains an integer T.(0<T<=20)
Then T test cases follow.
Each test case contains an interger N(1<=N<=25).It represents princess is N year(s) old.
For each test case,output the profile of the gift.
The uppercase letter of the level represents the jewelry box.
The first letter ‘Z‘ in ‘ZHI‘ represents the paper box.
2
1
2A AAAAA AZZZA AZBZA AZZZA AAAAA
1 #include <stdio.h> 2 #include <stdlib.h> 3 4 /* run this program using the console pauser or add your own getch, system("pause") or input loop */ 5 6 int main(int argc, char *argv[]) { 7 int t; 8 scanf("%d",&t); 9 10 if(t==1){ 11 printf("A"); 12 }else{ 13 char box[4*t-3][4*t-3]; 14 int j; 15 int k; 16 for(k=0;k<4*t-3;k++){ 17 box[0][k]=‘A‘; 18 } 19 20 for(k=1;k<4*t-4;k++){ 21 box[1][k]=‘Z‘; 22 } 23 box[1][0]=‘A‘; 24 box[1][4*t-4]=‘A‘; 25 26 int temp; 27 for(j=0;j<2*t-1;j++){//上半部分 28 if((j%2)==0&&j!=0){//偶数行 29 box[j][j-1]=‘Z‘; 30 box[j][(4*t-3)-j]=‘Z‘; 31 32 for(k=j;k<(4*t-3)-j;k++){ 33 box[j][k]=box[j-2][k]+1; 34 } 35 36 for(k=0;k<j;k++){ 37 if(k!=((4*t-3)-j)&&k!=j-1){ 38 box[j][k]=box[j-2][k]; 39 } 40 } 41 for(k=(4*t-3)-j;k<4*t-3;k++){ 42 if(k!=((4*t-3)-j)&&k!=j){ 43 box[j][k]=box[j-2][k]; 44 } 45 } 46 47 } 48 49 if((j%2)!=0&&j!=1){ 50 for(k=0;k<4*t-3;k++){ 51 if(k!=j-1&&k!=(4*t-3)-j){ 52 box[j][k]=box[j-2][k]; 53 }else{ 54 box[j][k]=‘A‘+(j-1)/2; 55 } 56 } 57 } 58 } 59 temp=2; 60 for(j=2*t-1;j<4*t-3;j++){ 61 for(k=0;k<4*t-3;k++){ 62 63 box[j][k]=box[j-temp][k]; 64 65 } 66 temp=2+temp; 67 } 68 69 70 for(k=0;k<4*t-3;k++){ 71 for(temp=0;temp<4*t-3;temp++){ 72 printf("%c",box[k][temp]); 73 printf("%s"," "); 74 } 75 printf("\n"); 76 } 77 78 } 79 return 0; 80 }
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原文地址：http://www.cnblogs.com/wpzy2311/p/5312494.html
