对于每对a.b,要求在一行输出a加b的和,每组数据输出占一行.

count==3时会打印两遍可以改成：if ( count == 3 ) {    System.o

小题1:去掉me小题2:去掉in小题3:usefully—useful小题4:that—what小题5:正确小题6:but—however小题7:did—do小题8:第二个it—them小题9:ver

1.troubles-trouble  2.geting-getting  3.the-a     4.&nb

1.bring改为brought  2.去掉with  3.happy改为happily  4.students改为student5.so改

1. chose-chosen 2.去掉at 3.about-with/to 4.abouthistory间加the 5.good-well6.was

#includevoidmain(){inti,N,sum;while(scanf("%d",&N)){sum=0;for(i=1;i

/*矩阵的行和列运行时输入*/#include#include#include#includevoidrp(int**a,intr,intc){inti,j;int**ta=(int**)mall

//分析：魔方阵有如下规律：//1：自然数1总是在方阵第一行当中一列上.//2：后续的自然数在当前数的右上方,//1）如果是在第一行则行数变为第n行列数加1；//2）如果是在最后一列,行数减1,列数为

#includevoidmain(){inta,b;while(1){scanf("%d%d",&a,&b);if(a==0&&b==0)break;printf("%d\n",a+b);}}

他没指定case个数.你输入n一点用也没有.这样就可以了.用下面的输入while(scanf("%d%d",&a,&b)!=EOF){printf("%d\n",a+b);}再问：什么叫没指定case

小题1:strong小题2:great小题3:practice小题4:also前加are小题5:去of小题6:√小题7:what小题8:learns小题9:himself小题10:good前the略

小题1:am—was小题1:在foreign前加a小题1:understand—understanding小题1:poorly—poor 小题1:He—She小题1:but—and 

小题1:a小题2:threw→throw小题3:listenMP3→listen∧toMP3小题4:students→students’小题5:good→well小题6:on→with小题7:hard

#includeintmain(){intx,y;while(scanf("%d%d",&x,&y)==2&&(x!=0||y!=0)){\x09printf("%d\n",x+y);}return0

16.Asailistoaboatwhatanengineittoacar.17.WhetherwewillvisittheGreatWalltomorrowhasnotbeendecided.18.

#include"stdio.h"#include"math.h"main(){inta,b,i,j,k,t,n=0;printf("请输入两个整数:");scanf("%d%d",&a,&b);if

#includeintmain(){printf("ThisisaCprogram\n");return0;}这个就可以了

假设你的数据放在变量a中.fori=1:32361[m,n]=find(a(i,:)>10);a(1:n(1)-1)=zeros(1,n(1)-1);ifsum(a(n,:))==3270a(n(1)

小题1:Get—Getting小题2:mostly—most小题3:found--founded小题4:inCanada∧-are小题5:rests-rest小题6:that–which小题7:pay

你只需要把intj=0改成staticj=0错误原因是,你在关于x的循环中每次调用f(x)的时候,j都重新置为0而实际上,j的值是需要保留的,如果置为static类型,在x增加的时候,j的值就不会重置