输入两个整数.若它们的平方和大于100,则
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#includevoidmain(){inta,b;printf("Inputtwointegers:");scanf("%d%d",&a,&b);printf("和:a+b=%d\n",a+b);p
#includevoidmach(intx,inty){printf("x²+y²=%d\n",(x*x+y*y));printf("x²-y²=%d\n",(
#includeintmain(){\x09inta,b;\x09scanf("%d%d",&a,&b);\x09printf("商为%.2f余数为%d\n",1.0*a/b,a%b);\x09ret
是输入结束不会还是取整数小数不会?再问:都不会,完全没有思路,只有寻求帮助了再答:������֪��˼·����ֱ��Ҫ����再问:ֱ��Ҫ����ϣ������C++��д��˳��˵������
programsample;vari,x,s,q:longint;begins:=0;q:=0;fori:=1to30dobeginread(x);inc(s,x);inc(q,sqr(x));end
#includevoidmain(){floata,b,sum,change;scanf("%f,%f",&a,&b);sum=a*a+b*b;if(sum>100){change=sum/100;w
#include#incluedvoidmain(){inta;intb;intc,d;a=1500,b=350;c=a/b;//商数d=a%b;//余数pritf("商数为:%d\n",c);pri
#includeintmain(){inta,b,c,d=0;scanf("%d%d",&a,&b);c=a/b;d=a%b;printf("商是:%d\n",c);printf("余数是:%d\n"
#include"stdio.h"main(){inta=0,b=0,c=0,d=0;printf("请输入两个整数:");scanf("%d,%d",&a,&b);if(b==0)printf("输
#includevoidmain(){inta,b;scanf("%d%d",&a,&b);printf("%d",a*a+b*b);}
http://zhidao.baidu.com/link?url=wyic3NEB45i2K9YZE2SpNiAV4aD7rZHpluBlyF76zg0n_8wr1t6-KOYm6eYY23KqG5n
设连续整数分别是n和n1.其和是2n1.其平方和是n^2(n1)^2=2n^22n1.因为n和n1互质,所以n和n(n1)=2n1互质,2n1是单数,所以2×n×n=2n^2和2n1互质.所以2n1和
#include#defineDEX(x,y)(x*x+y*y)intmain(void){inta,b;printf("输入两个数:");scanf("%d%d",&a,&b);printf("结果
726=2×3×11×11所以726=22×33这两个整数是22和33
#include#defineSURPLUS(a,b)(a%b)main(){inta,b;printf("pleaseinputinteger:");scanf("%d,%d",&a,&b);pri
设这五个连续整数为n,n+1,n+2,n+3,n+4,n2+(n+1)2+(n+2)2=(n+3)2+(n+4)2,解得n=10或n=-2,当n=10时这五个数为10,11,12,13,14,当n=-
x^2+(x+1)^2+(x+2)^2=(x+3)^2+(x+4)^2x^2-8x-20=0x=1010,11,12,13,14
#includevoidmain(){inta,b,c,min,max,cha;scanf("%d%d%d",&a,&b,&c);min=a;max=a;if(b>max){max=b;}if(bma
//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){intm,n,k,l;while(1){printf("Typem
设m可以表示为两个整数的平方和,即有整数a,b,使m=a^2+b^2则2m=2(a^2+b^2)=a^2+b^2+a^2+b^2=a^2+2ab+b^2+a^2-2ab+b^2=(a+b)^2+(a-