4、for (a=1,b=1;a=20) break;if(b%3==1){b+=3; continue;}b-=5; } 程序的输出结果a的值为__________ .
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4、for(a=1,b=1;a=20)break;if(b%3==1){b+=3;continue;}b-=5;}程序的输出结果a的值为__________.4、for(a=1,b=1;a=20)br
4、for (a=1,b=1;a=20) break;if(b%3==1){b+=3; continue;}b-=5; } 程序的输出结果a的值为__________ .
4、for (a=1,b=1;a=20) break;
if(b%3==1)
{b+=3; continue;}
b-=5; }
程序的输出结果a的值为__________ .
4、for (a=1,b=1;a=20) break;if(b%3==1){b+=3; continue;}b-=5; } 程序的输出结果a的值为__________ .
a=8
int a,b; for(a=1;a
for什么时候加{}花括号#includemain(){int a,b=1;for(a=1;a
for循环题目int a,b;for(a=1;a
main() {int a,b; for (a=1,b=1;a
vb for a=0 for i=1 to 10 a=a+1 b=0 for j=1 to 10 a=a+1 b=b+2 next j next i print a,b 结果
If =a(a+1)/2 for all integres a and b =,then is( )
有以下程序,#include Main(){char A,B,C;B=’1’;C=’A’For(A=0;A
#include Main() {char A,B,C; B=’1’;C=’A’ for(A=0;A
#include void main() { int a,b,c; for(a=1; a
/bin/bash for ((a=1;a
#include int main(){int a,b,c; int i = 0; for( c = 1; c = 33; ++ c ) for( b = 100; b = c; -- b )for( a = 100; a = b; -- a) { if( (a + b + c < 100) && ( (a * a * b * b) == ((a * a + b * b) * c * c) )++ i; } printf(符合条件的组数为:%d
,i);
#include int main(){int a,b,c; int i = 0; for( c = 1; c = 33; ++ c ) for( b = 100; b = c; -- b )for( a = 100; a = b; -- a) { if( (a + b + c < 100) && ( (a * a * b * b) == ((a * a + b * b) * c * c) )++ i; } printf(符合条件的组数为:%d
,i);
C语言(char*a char*b)*a=*a^*b;*b=*a^*b;*a=*a^*b;当a b为同一个数的时候 为什么最后a b都变为null了void swap(char *a,char *b){ *a=*a^*b;*b=*a^*b;*a=*a^*b;}int main(){char a[5]={'1','2','3','4','5'};for(int i=0,j=4;i
A is based on B=B is ( ) for A
For a = 1 To 100 For b = 1 To 100 For c = 1 To 100 If a ^ 2 + b ^ 2 = c ^ 2
main( ){int a,b;for(a=1,b=1;a=20)break;if(b%3==1){b+=3;continue;}b-=5;}printf(a=%d
,a);}8
请问matlab怎么改进for循环比如说for a=1:100for b=1:100c=(a*b)+(a/b);endend这个循环好费时间啊~请问怎么改进下不用for循环~
关于lingo程序的,sets:A/1..4/:c,b,n;X/A,A/:D;endsetsmin=1.1*n(1)+1.2*n(2)+1.3*n(3)+1.4*n(4);@for(A(i):@sum(A(j):c(i)*D(i,j)=1750);@for(A(i):@sum(A(j):n(i)*D(i,j)=b(i));@for(A(i):@gin(n(i)));@for(A(j):@for(A(i):@gin(D(i,j))));n(1)>n(2);n(2)>n(3);n