题目链接http://acm.hdu.edu.cn/showproblem.php?pid=2504题目描述思路就是写个gcd函数直接调用就行了。直接看代码吧ac代码#include#defineN100000intgcd(inta,intb){returnb==0?a:gcd(b,a%b);}intmain(){intn,m;inti;intt;scanf("%d",&t);while(t--)

代码一：//这个没有过#include//typedeflonglongll;intmain(){intT;scanf("%d",&T);longlonga,b;while(T--){scanf("%lld%lld",&a,&b);for(longlongi=1+b;i//typedeflonglongll;intgcd(inta,intb){if(a#include#include#includ

思路gcd（a/b,c/b）==1#include#include#includeusingnamespacestd;typedeflonglongll;intn,m;intgcd(inta,intb){if(b==0)returna;returngcd(b,a%b);}intmain(){inti,j,k,t;inta,b;cin>>t;while(t--){cin>>a>>b;inttmp=a

#includeusingnamespacestd;intgcd(inta,intb){intt,r;if(a>b){t=b;b=a;a=t;}while(a){r=b%a;b=a;a=r;}returnb;}intmain(){intT;cin>>T;//gcd(a,c)==b;//c==b*ninta,b,c;while(T--){cin>>a>>b;for(c=b+b;gcd(a,c)!

ProblemDescription有三个正整数a,b,c(0 usingnamespacestd; intyue(intm,intn) { inttmp; while(n) { tmp=m; m=n; n=tmp%n; } returnm; } intmain() { intn,a,b,c; cin>>n; while(n--) { cin>>a>>

又见GCDTimeLimit:1000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):9984    AcceptedSubmission(s):4157ProblemDescription有三个正整数a,b,c(0 intgcd(inta,intb){ returnb?gcd(b,a

HDU_Steps2.1解题2.1.1 HDU1108最小公倍数如题,a*b/gcd(a,b)即可2.1.2 HDU2138HowManyPrimesNumbers直接判断会超时,直接打表空间不够,于是采用了部分打表,部分判断的方法.似乎有一种随机素数测试的方法做这一题会更快一些,但没有上述方法简单#include #include #include #include #include #incl