1#import"Common.h"23staticNSArray *arrPartOfURL =nil;4@implementationCommon56+ (NSURL *)randomImageURL {7if(arrPartOfURL ==nil) {8arrPartOfURL =@[9@"chronograph",10@"color",11@"modular",12@"utility",13@"mickey_mouse",14@"simple",15@"motion",16@"solar",17@"astronomy",18@"ti...
63 { 64 [self gcdRequest]; 65 break; 66 } 67 68 default: 69 break; 70 } 71 } 72 73 #pragma mark - 74 #pragma mark GCDRequest 75 76 - (void) gcdRequest 77 { 78 GCDHelper *hp = [GCDHelper new]; 79 80 [hp gcdHttpRequestWithURL:@"http://localhost:8888/test.php" 81 http...
1#import"Common.h"23staticNSArray *arrPartOfURL =nil;4@implementationCommon56+ (NSURL *)randomImageURL {7if(arrPartOfURL ==nil) {8arrPartOfURL =@[9@"chronograph",10@"color",11@"modular",12@"utility",13@"mickey_mouse",14@"simple",15@"motion",16@"solar",17@"astronomy",18@"ti...
is the smallest positive integer that is divisible by both and . For example, Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair? Input The first and only line contains an integer Output Print two positive integers, and , such that the value ...
GCD - Extreme (II) UVA - 11426 数学 N , you will have to nd the value of G . The de nition of G is given below: G = i<N ∑ i =1 j N ∑ j = i +1 GCD ( i; j ) Here GCD ( i; j ) means the greatest common divisor of integer i and integer j . For those who ...
HDU - 5381 - The sum of gcd 左端点相同的区间最多有 log(max(ai)) 个gcd ,且随着右端点的变大而变小。右端点相同的区间同理。 于是考虑 O(nlogn) 预处理出来。 然后用莫队维护区间 gcd 和就行了。 #include<bits/stdc++.h> #define fi first #define se second #define mp make_pair ...
The input contains multiple test cases, each of which contains two positive integers, the GCD and the LCM. You can assume that these two numbers are both less than 2^63. Output For each test case, output a and b in ascending order. If there are multiple solutions, output the pair with...
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[4000008]; int inof[4000008]; int deep,cnt; H() { cnt=0; deep=0; memset(inof,-1,sizeof inof); } void add(int x,int key,int S) { A[cnt]=S; E[deep]=point(key,cnt++,inof[x]); inof[x]=deep++; } int find(int key) { int d=key%4000007; for(int i=inof[d];i...