C++程序题.doc

假定：下面程序中均已经包含相应的头文件 #include ，等。1 void main() int a= -1, b=3, k1, k2; k1=(a+0) & (!(b-=0); k2=(-a0) & (!(-b=0);cout”a=”a” b=”bendl;cout”k1=”k1” k2=”k2endl;2 void main() int n=0,x=39582, t=x; while(t) coutt%10; t=t/10; n=n+1; coutendlx”是一个：”n”位数字。”endl;3 void main() int a=1, b=2, c=5; c+=a+b+4; cout”a=”a” b=”b” c=”cendl; c=(a+, a+=b, a+b); cout”a=”a” b=”b” c=”cendl;4 void main() int x=3, y=6, z=0; while(x+!=(y=y-1) z+; if(yx) break; cout”x=”x”, y=”y”, z=”zendl;5 void main() int *p, i, j, a34; for(i=0;i3;i+) for(j=0;j4;j+) aij=2*i+3*j+1; for(p=&a00, j=1; j13; j+) cout*p” “; p+;if(j%4=0) coutendl; coutendl;6 int prem(int x)int k, t=sqrt(x);for(k=2; kt) return 1; else return 0; void main() for(int k=17; k33; k+=2)if(prem(k) coutk”, “; coutendl;7 void main() int m, n, m1, n1, t, max, min;coutmn; m1=m; n1=n; while(n1) t=m1%n1; m1=n1; n1=t; max=m1; min=m*n/max; coutm”和”n”的最大公约数是：”max”最小公倍数是：”minendl;8 void main() int z1=2, z2=3, m1=1, m2=2, k; for(k=1; k5; k+) coutz1/m1”, “z2/m2”, “;z1=z1+z2; z2=z1+z2;m1=m1+m2; m2=m1+m2;9 void main() int i, j, n=20, h=5; for(i=1; ih; i+) for(j=1; jn-2*i; j+)printf(“%2c” , ); for(j=1; j=2*i-1; j+)printf(“%2c”, *); printf(“n”); 10 void main() int k, j, h=6; for (k=1; kh; k+) /h行循环输出for(j=1; j=h-k; j+) cout” “; /输出3空格for(j=1; j=k; j+) coutsetw(3)=1; j-) coutsetw(3)j; coutendl; 11 void main() char st80; int count(char a); cout”请输入一行任意字符，回车换行结束：”; cin.getLine(st,80); /获取小于80个字符的键盘输入行。输入：a123b4?&f789g69# cout”count=”count(st)=A & ak=a & ak=z) num+;return num;12 void sort(int *p, int n) int i, j, k, t; for(i=0; in-1; i+)k=i;for(j=i+1; j*(p+k) k=j;if(k!=i) t=*(p+i); *(p+i)=*(p+k); *(p+k)=t; void main() int k,a5=59, 38, 29, 81,47; sort(a,5); for(k=0; k5; k+) coutak” “; coutendl;13 void main() int k, z1, z2, m1, m2;z1=2; z2=3; m1=1; m2=2; for(k=0; k10; k+)coutz1”/”m1”, ”z2”/”m2”, “;if(k+1)%3=0) coutendl;z1=z1+z2; z2=z1+z2; m1=m1+m2; m2=m1+m2; 14 void main() coutx; t=x; while(t) coutt%10; t=t/10; n=n+1; coutendlx”是一个：”n”位数字。”endl;15 void main() int a34=1,2,3,4,5,6,7,8,9,10,11,12; int i, j, *p=&a00, k2=0; for(i=0; i3; i+) for(j=0; j4; j+) k*p%3=0+=*p; p=p+1; couut”k0=”k0”, k1=”k1x; /由键盘输入数字9 for(i=0; i10; i+) if(ai!=x) a+k=ai; for(i=0; i=k; i+) coutai” “; couutendl;17 void mai