春C语言程序设计上答案.pdf

1 想翔 2017 春 C 语言程序设计 上（宣城校区） 第二周 2 想翔 3 想翔 第三周 1.让我们学习用 C 语言向世界问好。 #include int main() printf(“Hello world.n“); return 0; 2. 学学 scanf 和 printf 的用法吧 #include #define PAI 3.1415926 int main() int r,h; double s,v; scanf(“%d%d“, s=2*PAI*r*h; v=PAI*r*r*h; printf(“s=%.2f,v=%.2fn“,s,v); return 0; 3.输出字符的 ASCII 码 #include int main() char c; scanf(“%s“, printf(“The ASCII of character %c is %d.“,c,c); return 0; 4 想翔 4.一年级的算术题 #include int main() int x,y; scanf(“%d%d“, printf(“%d+%d=%dn“,x,y,x+y); printf(“%d-%d=%dn“,x,y,x-y); printf(“%d*%d=%dn“,x,y,x*y); printf(“%d/%d=%dn“,x,y,x/y); printf(“%d%d=%dn“,x,y,x%y); return 0; 5 判断三角形的形状 #include #include int main() int a,b,c; scanf(“%d%d%d“, if(a+bc else if(a=b | b=c | c=a) printf(“isoceles triangle.n“); else printf(“triangle.n“); else printf(“non-triangle.n“); return 0; 5 想翔 第四周 6 想翔 1. 计算时钟的夹角 #include int main() float m,d; int h,p; scanf(“%d%f“, p=h; if(p=12) p=0; if(p*30+m/2)m*6) d=p*30+m/2-m*6; else d=m*6-p*30-m/2; if(m 30) printf(“day is error.n“); return 0; break; case 2:if(year % 4 = 0 return 0; else if(day 28) printf(“day is error.n“); return 0; break; if(year % 4 = 0 for(n = 0;n 999 n = maxnum; n+) if(n % 2 = 0) q = n / 1000; b = n % 1000 / 100; s = n % 100 / 10; g = n % 10; if(q != b printf(“%d “,n); printf(“n“); printf(“counter=%dn“,count); count = 0; else printf(“Errorn“); 11 想翔 return 0; 3.求最小 m 值 #include int main() int max, i = 0, sum = 0; scanf(“%d“, while(sum max) i+; sum += i; printf(“%dn“,i); 第六周 12 想翔 13 想翔 1.数字正方形 #include int main() int n, x, y, count = 0; scanf(“%d“, for(y = 0; y n; y+) for(x = 0; x n; x+) count+; printf(“%3d“,count); printf(“n“); return 0; 2.空心型字符 #include int main() int n, i, q, p; char ch; scanf(“%c%d“, 14 想翔 for(i = 0; i n; i+) for(q = i; q n - 1; q+) printf(“ “); printf(“%c“,ch + i); for(q = 0; q = 0; i-) for(q = i; q 0; q-) printf(“ “); if(q = 1) printf(“%c“, ch + i); printf(“n“); return 0; 3.邮票组合 #include int main() int i5,t3126,n,n1,n2,n3,n4,temp = 0; i0 = 0; for(n = 1; n 5; n+) scanf(“%d“, for(n = 0; n 3126; n+) tn = 0; for(n = 0; n 5; n+) for(n1 = 0; n1 5; n1+) 15 想翔 for(n2 = 0; n2 5; n2+) for(n3 = 0; n3 5; n3+) for(n4 = 0; n4 5; n4+) ttemp = in + in1 + in2 + in3 + in4; temp+; for(n = 0; n 3124; n+) for(n1 = n; n1 tn1) temp = tn; tn = tn1; tn1 = temp; temp = 0; for(n = 0; n 3125; n+) if(tn tn + 1 ) temp+; if(temp != tn + 1) break; printf(“The max is %d.n“,temp - 1); return 0; 第七周 1.组成最大数 16 想翔 #include int main() char s50,t; int n = 0, i, m; scanf(“%s“,s); while(sn != 0) n+; for(i = 0; i n - 1; i+) for(m = i + 1; m n; m+) if(si sm) t = si; si = sm; sm = t; printf(“%sn“,s); 2.文回数 #include int main() int n, m, square, i, temp, log; char s6; scanf(“%d“, for(m = 1; m n; m+) i = 0; square = m * m; log = 1; while(square != 0) temp = square % 10; si = 0 + temp; i+; square /= 10; 17 想翔 for(temp = 0; temp i / 2; temp+) if(stemp != si - 1 - temp) log = 0; if(log) printf(“%dn“,m); return 0; 3.洗牌 #include int main() int a10000,tmp15000,tmp25000, i, m, n = 0, log = 0, total = 0; scanf(“%d“, / for(m = 0; m 2* i; m+) am = m; do for(m = 0; m i; m+) tmp1m = am; tmp2m = am + i; n = 0; for(m = 0; m i; m+) an = tmp2m; n+; an = tmp1m; n+; log = 0; for(m = 0; m 2 * i; m+) if(am != m) log = 0; break; 18 想翔 if(am = m) log = 1; total+; while(log != 1); printf(“%dn“,total); return 0; 4.字数整除 #include main() int k, a,num1,num2,num3; scanf(“%d“, for (a=10000;a!=30001;a+) num1=a/100; num2=a/10-(a/10000)*1000; num3=a-(a/1000)*1000; if (num1=0|num2=0|num3=0) continue; else if(num1 % k)=0 getchar(); getchar(); return 0; 第八周 1.单词排序 #include #include int main() 19 想翔 char s5100, t100; int i, n; for(i = 0; i 5; i+) scanf(“%s“,si); for(i = 0; i 4; i+) for(n = i + 1; n 5; n+) if(strcmp(si,sn) 0) strcpy(t,si); strcpy(si,sn); strcpy(sn,t); for(i = 0; i 5; i+) printf(“%sn“,si); return 0; 2.寻找矩阵的鞍点 #include #include int main() int i100100, m, n, x, y, temp, log = 0; scanf(“%d%d“, for(y = 0; y m; y+) for(x = 0; x n; x+) scanf(“%d“, 20 想翔 /将每行的最大值放入每行的最后； for(y = 0; y m; y+) temp = 0; for(x = 0; x temp) temp = iyx; iyn = temp; /将每列的最小值放入每列的最后 for(x = 0; x n; x+) temp = 0x7fffffff; for(y = 0; y m; y+) if(iyx temp) temp = iyx; imx = temp; for(y = 0; y m; y+) for(x = 0; x n; x+) if(iyx = imx) log = 1; if(log = 0) printf(“No Pointn“); return 0; 21 想翔 1. 身份证的奥秘 #include #include int main() int n, i, m, j, sumt, y; char s10019, temp19, t; scanf(“%d“, for(i = 0; i n; i+) scanf(“%s“,si); for(i = 0; i n; i+) m = 0; while(sim != 0) m+; if( m != 15 if( m = 15) for(j = 0; j 5) temp6 = 1; temp7 = 8; else temp6 = 1; temp7 = 9; for(j = 6; j 16; j+) tempj + 2 = sij; strcpy(si,temp); sumt = (si0 - 0) * 7 + (si1- 0) * 9 + (si2 - 0) * 10 + (si3 - 0) * 5 + (si4 - 0) * 8 + (si5 - 0) * 4 + (si6-0) * 2 + (si7-0) * 1 + (si8-0) * 6 + (si9-0) * 3 + (si10-0) * 7 + (si11-0) * 9 + (si12-0) * 10 + (si13-0) * 5 + (si14-0) * 8 + (si15-0) * 4 + (si16-0) * 2; /sumt = atoi(si0) * 7 + atoi(si1) * 9 + atoi(si2) * 10 + atoi(si3) * 5 +atoi(si4) * 8 + atoi(si5) * 4 + atoi(si6) * 2 + atoi(si7) * 1 + atoi(si8) * 6 + atoi(si9) * 3 + atoi(si10) * 7 + atoi(si11) * 9 + 22 想翔 atoi(si12) * 10 + atoi(si13) * 5 + atoi(si14) * 8 + atoi(si15) * 4 + atoi(si16) * 2; y = sumt % 11; switch (y) case 0: si17 = 1; break; case 1: si17 = 0; break; case 2: si17 = X; break; case 3: si17 = 9; break; case 4: si17 = 8; break; case 5: si17 = 7; break; case 6: si17 = 6; break; case 7: si17 = 5; break; case 8: si17 = 4; break; case 9: si17 = 3; break; case 10: si17 = 2; break; si18 = 0; printf(“%sn“,si); if(m = 18) sumt = (si0 - 0) * 7 + (si1- 0) * 9 + (si2 - 0) * 10 + (si3 - 0) * 5 + (si4 - 0) * 8 + (si5 - 0) * 4 + (si6-0) * 2 + (si7-0) * 1 + (si8-0) * 6 + (si9-0) * 3 + (si10-0) * 7 + (si11-0) * 9