while(scanf("%d",&m)!=EOF)

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while(scanf("%d",&m)!=EOF)while(scanf("%d",&m)!=EOF)while(scanf("%d",&m)!=EOF)我想实现多组数据的循环输入,每组数据包含2个

while(scanf("%d",&m)!=EOF)
while(scanf("%d",&m)!=EOF)

while(scanf("%d",&m)!=EOF)
我想实现多组数据的循环输入,每组数据包含2个数,中间用一个空格隔开.
最后我想输入ctrl + Z 然后回车 来结束.
如果我用while(scanf("%d %d",&a,&b)!=EOF)来做的话,就无法通过输入ctrl + Z 然后回车 来结束.
scanf ()的返回值是实际得到的数据的个数 如果没有任何字符成功读入 scanf的返回值是EOF
这里,你按的回车也是字符被读入了,所以退不出

while(scanf(%d,&m)!=EOF) while (scanf(%d%d, &m, &n))什么意思? #include < stdio.h > main() { int m,n; scanf(%d%d,&m,&n); while (m!=n) { while(m>n) m=m-n;while(m while(scanf(%d%d,&a,&b)! while (scanf(%d/%d,&i,&j), 请问在c语言里:scanf(%d,&m);while(i main () {int m,n; scanf (%d%d,&m,&n); while (m!=n) { while(m>n) m-=n; while(n>m) m-=m; }printf(m=%d ,m)}如果键盘上输入65 14 结果是? # include # include int mian () { int m,n,i,s=0; int a[10000]; while(scanf(%d, for (scanf(%d,&m);m;m--)是什么意思 while(scanf(%d,&n)&&n!=-1)和while(scanf(%d,&n)!=-1)有什么区别》》具体的解释 #include int main() { int a[150];int m,n,i,j,t; while(scanf(%d %d,&m ,&n)!=EOF) {printf(#includeint main(){int a[150];int m,n,i,j,t;while(scanf(%d %d,&m ,&n)!=EOF){printf(%d~%d prime include:,m,n);t=0;for ( i=m+1;i c语言 集合a-b#include #include int cmp(const int *a,const int *b){return *a - *b;}int main(void){int n,m,i,j;int s[101];while (scanf(%d%d,&n,&m),m+n){for (i = 0; i < n; i++)scanf(%d,s + i);for (i = 0; i < m; i++){scanf(%d,s + n);for (j = 0; 一个简单编程题#include#includeint main(){int n,i=1,m=1;double x,s=0;scanf(%d%d,&x,&n);while(i 关于C语言:while(scanf(%d%d%d,&n,&k,&m)!=EOF) 求精确解释这个循环的含义,尤其是关于“!=EOF”.前面只接触过类似while(n 关于C语言:while(scanf(%d%d%d,&n,&k,&m)!=EOF) 求精确解释这个循环的含义,尤其是关于“!=EOF”.前面只接触过类似while(n #include int ZDGYS(int m,int n) { int r; scanf(%d%d,&m,&n); r=m%n; while(r!=0) { m=n; n=#includeint ZDGYS(int m,int n){int r;scanf(%d%d,&m,&n);r=m%n;while(r!=0){m=n;n=r;r=m%n;}return n;}main(){printf(%d,r);}求m,n的最大公约数 看看哪有 scanf(%d, 杭电acm2034求解释#include main(){ int a[100] , b[100] , i , j , n , m , f , t ; while((scanf(%d%d , &i , &j) != EOF) && i != 0 && j != 0) { for(n = 0 ; n < i ; n++) { scanf(%d , &a[n]); } for(n = 0 ; n < j ; n++) { scanf(%d ,