若有以下程序段,则数值为3的表达式( )。int x[12]={1,2,3,4,5,6,7,8,9,10};char c=‘a’;char d='d',f;
A. x[d-c]
B. x['d'-'b']
C. x[3]
D. x['d'-c]
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执行下面的程序片段,k的值是() int k=0,i,j; for(i=0;i<5;i++) for(j=0;j<3;j++) k=k+1;
A. 5
B. 15
C. 3
D. 8
下面程序的运行结果是( )。#include int main( ){ int a[10],i,k=1; for(i=0;i<10;i++) a[i]=i; for (i=1;i<6;i++) k+=a[i]+i; printf("%d\n",k); return 0;}
A. 20
B. 32
C. 31
D. 33
现有如下程序,则程序的输出结果为( )。#include int main( ){ int k[30]={12,324,45,6,768,98,21,34,453,456}; int count=0,i=0; while(k[i]) { if(k[i]%2==0‖k[i]%5==0) count++; i++; } printf("%d,%d\n",count,i); return 0;}
A. 7,10
B. 8,10
C. 8,8
D. 7,8
以下程序段( )不能实现求s = 1+2+...+n-1。
A. int i, n, s = 0; scanf("%d", &n);for ( i = 1; i < n; i++ ) { s = s + i; }
B. int i, n, s = 0; scanf("%d", &n);for (i = n-1; i > 0; i-- ) { s = s + i; }
C. int i, n, s = 0; scanf("%d", &n);for ( i = n-1; i > 0; ++i ) { s = s + i; }
D. int i, n, s = 0; scanf("%d", &n);for ( i = 1; i <= n-1; ++i ) { s = s + i; }