写完这道题目才发现自己对二分的理解太浅了 这题是典型的利用二分“假定一个问题可行并求最优解” 二分是通过不断缩小区间来缩小解的范围,最终得出解的算法 我们定义一个c(x) 表示判断函数 如果对任意y>=x 当x满足条件的时候 y也满足条件 那么我们就一个不断缩小区间范围来确定最后的解 好扯了这么多犊子 来说下这道题目.. 我们的目的是对A,B 有 (x+A)/(y+B) == (p/q) 其中A<=B 那么可以转换为 A+x=k*p  B+y=k*q; 既 A=k*p-x, B=k*q-y;(…

You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissions, out of which x have been successful. Thus, your current success rate on Codeforces is equal to x / y. Your favorite ratio…

C. Success Rate time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made ysubmissio…

C. Success Rate time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are an experienced Codeforces user. Today you found out that during your activity on Codeforces you have made y submissi…

http://codeforces.com/problemset/problem/807/C C. Success Rate time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You are an experienced Codeforces user. Today you found out that during your…

Success Rate 思路: 水题: 代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define ll long long inline void in(ll &now) { ; ') Cget=getchar(); ') { now=now*+Cget-'; Cget=getchar(…

Success Rate CodeForces - 807C 给你4个数字 x y p q ,要求让你求最小的非负整数b,使得 (x+a)/(y+b)==p/q,同时a为一个整数且0<=a<=b. (0 ≤ x ≤ y ≤ 109; 0 ≤ p ≤ q ≤ 109; y > 0; q > 0) 解法: (x+a)/(y+b)==p/q; --> x+a=np; y+b=nq; --> a=np-x; b=nq-y; --> 二分n; #include <cs…

Success Rate 题目链接 题意 给你两个分数形式的数,然后有两种变化方式 上下都+1 仅下面部分+1 让你求第一个分数变化到第二个分数的最小步数. 思路 有几种特殊情况分类讨论一下. 首先我们先把右边化为最简比. 我们可以得到方程$$\frac{(x+a)}{(y+b)} = \frac{kp}{kq}$$ \[a = k*p - x\\b = k*q - y \] 根据\(b >= a\)可求得\(k = \frac{y-x}{q-p}\)向上取整,并且要求a>=0: 代码 #in…

#include<cstdio> #include<iostream> #include<string> #include<algorithm> #include<iterator> using namespace std; /* * A solution for "The Trip" problem. * UVa ID: 10137 */ #include <stdio.h> int main (int argc…

#include<cstdio> #include<iostream> #include<string> #include<algorithm> #include<iterator> #include<cstring> using namespace std; int n, m; ][]; void calc() { ;i<n;i++) { ;j<m;j++) { if(x[i][j]=='*') { putchar('*…