模拟
                            
                            
                            問(wèn)題描述  
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..$$$$$$$$$$$$$.. 提示 請(qǐng)仔細(xì)觀察樣例,尤其要注意句點(diǎn)的數(shù)量和輸出位置。 傻逼模擬題 char ans1[9][10] = {"..$$$$$..","..$...$..","$$$.$.$$$","$...$...$","$.$$$$$.$","$...$...$","$$$.$.$$$","..$...$..","..$$$$$.." }; struct ans{char str[140][140]; }aa[30]; void OUT(int n){rep(i,0,9) rep(j,0,9){aa[1].str[i][j] = ans1[i][j];}repf(i,2,n){rep(j,0,9+4*(i-1)) rep(k,0,9+4*(i-1)) aa[i].str[j][k] = '.';rep(j,2,2+9+4*(i-2)){rep(k,2,2+9+4*(i-2)){aa[i].str[j][k] = aa[i-1].str[j-2][k-2];}}rep(j,2,2+9+4*(i-2)) aa[i].str[0][j] = '$';///aa[i].str[1][2] = '$';aa[i].str[1][2+9+4*(i-2)-1] = '$';//rep(j,2,2+9+4*(i-2)) aa[i].str[2+9+4*(i-2)+1][j] = '$';///aa[i].str[2+9+4*(i-2)][2] = '$';aa[i].str[2+9+4*(i-2)][2+9+4*(i-2)-1] = '$';///rep(j,3,2-1+9+4*(i-2)) aa[i].str[j][2+1+9+4*(i-2)] = '$',aa[i].str[j][2+9+4*(i-2)] = '.',aa[i].str[j][0] = '$',aa[i].str[j][1] = '.';aa[i].str[2][0] = '$',aa[i].str[2][1] = '$';aa[i].str[2][2] = '$';aa[i].str[2+9+4*(i-2)-1][0] = '$',aa[i].str[2+9+4*(i-2)-1][1] = '$';aa[i].str[2+9+4*(i-2)-1][2] = '$';aa[i].str[2][2+9+4*(i-2)-1] = '$';aa[i].str[2][2+9+4*(i-2)] = '$';aa[i].str[2][2+9+4*(i-2)+1] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)-1] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)+1] = '$';}rep(i,0,9+4*(n-1)) printf("%s\n",aa[n].str[i]); } int main() {//freopen("in.txt","r",stdin);int n;while(cin>>n){OUT(n);}return 0; } View Code
                        小明為某機(jī)構(gòu)設(shè)計(jì)了一個(gè)十字型的徽標(biāo)(并非紅十字會(huì)啊),如下所示:
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對(duì)方同時(shí)也需要在電腦dos窗口中以字符的形式輸出該標(biāo)志,并能任意控制層數(shù)。
輸入格式 一個(gè)正整數(shù) n (n<30) 表示要求打印圖形的層數(shù)。 輸出格式 對(duì)應(yīng)包圍層數(shù)的該標(biāo)志。 樣例輸入1 1 樣例輸出1 ..$$$$$....$...$..
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..$$$$$$$$$$$$$.. 提示 請(qǐng)仔細(xì)觀察樣例,尤其要注意句點(diǎn)的數(shù)量和輸出位置。 傻逼模擬題 char ans1[9][10] = {"..$$$$$..","..$...$..","$$$.$.$$$","$...$...$","$.$$$$$.$","$...$...$","$$$.$.$$$","..$...$..","..$$$$$.." }; struct ans{char str[140][140]; }aa[30]; void OUT(int n){rep(i,0,9) rep(j,0,9){aa[1].str[i][j] = ans1[i][j];}repf(i,2,n){rep(j,0,9+4*(i-1)) rep(k,0,9+4*(i-1)) aa[i].str[j][k] = '.';rep(j,2,2+9+4*(i-2)){rep(k,2,2+9+4*(i-2)){aa[i].str[j][k] = aa[i-1].str[j-2][k-2];}}rep(j,2,2+9+4*(i-2)) aa[i].str[0][j] = '$';///aa[i].str[1][2] = '$';aa[i].str[1][2+9+4*(i-2)-1] = '$';//rep(j,2,2+9+4*(i-2)) aa[i].str[2+9+4*(i-2)+1][j] = '$';///aa[i].str[2+9+4*(i-2)][2] = '$';aa[i].str[2+9+4*(i-2)][2+9+4*(i-2)-1] = '$';///rep(j,3,2-1+9+4*(i-2)) aa[i].str[j][2+1+9+4*(i-2)] = '$',aa[i].str[j][2+9+4*(i-2)] = '.',aa[i].str[j][0] = '$',aa[i].str[j][1] = '.';aa[i].str[2][0] = '$',aa[i].str[2][1] = '$';aa[i].str[2][2] = '$';aa[i].str[2+9+4*(i-2)-1][0] = '$',aa[i].str[2+9+4*(i-2)-1][1] = '$';aa[i].str[2+9+4*(i-2)-1][2] = '$';aa[i].str[2][2+9+4*(i-2)-1] = '$';aa[i].str[2][2+9+4*(i-2)] = '$';aa[i].str[2][2+9+4*(i-2)+1] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)-1] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)] = '$';aa[i].str[2+9+4*(i-2)-1][2+9+4*(i-2)+1] = '$';}rep(i,0,9+4*(n-1)) printf("%s\n",aa[n].str[i]); } int main() {//freopen("in.txt","r",stdin);int n;while(cin>>n){OUT(n);}return 0; } View Code
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轉(zhuǎn)載于:https://www.cnblogs.com/DreamHighWithMe/p/3565391.html
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