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.
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count==3时会打印两遍可以改成:if ( count == 3 ) { System.o
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1.bring改为brought 2.去掉with 3.happy改为happily 4.students改为student5.so改
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#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
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#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)
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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