三 0 (mod mj), i 孝j, 1 <i, j <no
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1.三 0 (mod mj), i 孝j, 1
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2.for(i=0;i
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3.for (i=0; i
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4.int i, j, x=0;Eor (i=l; i
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5.程序设计(第28套)void fun ( int m, int *a , int *n ){int i,j=0;for(i=1;i<=m;i )if(i%7==0||i%11==0)A、[j ]
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6.int m=0, i, j;for(i=l;i<=n;i )for(j=1;j<=2 * i;j )m ;
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7.int i, j, x=0;Eor (i=l; i(n; i )for (j=i l; i<=n; j X ;
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8.#include void main(){ int a[6]={12,4,17,25,27,16},b[6]={27,13,4,25,23,16},i,j;for(i=0;i〈6;i ) {for
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9.循环 for(i=0, j=5; ++i!=--j; ) printf(“%d %d”, i, j); 将执行_
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10.int i=0,j=0,a=6;if ((++i>0) || (++j>0 )) a++ ;System.out.println(“i=”+i+” ,j=”+j+” ,a=”+a);