摘要：
http://www.acmore.net/problem.php?id=1467根据朴素的欧几里德原理有 gcd(a,b)=gcd(b,a mod b);则:ax1+by1=bx2+(a mod b)y2;即:ax1+by1=bx2+(a-[a/b]*b)y2=ay2+bx2-(a/b)*by2;根据恒等定理得：x1=y2; y1=x2-[a/b]*y2;本题直接利用这个结论。#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include < 阅读全文
摘要：
二分+树状数组动态的寻找比a大k的数#include <stdio.h>#include <iostream>#include <string>#include <string.h>#include <algorithm>#include <stdlib.h>#include <math.h>#include <vector>#include <map>using namespace std;const int maxn = 100005;int c[maxn];int lowbit(i 阅读全文