B. Robin Hood 题目连接: http://www.codeforces.com/contest/671/problem/B Description We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor. There are n citiz…

D. Robin Hood   We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor. There are n citizens in Kekoland, each person has ci coins. Each day, Robin Hood…

B. Robin Hood time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, a…

B. Robin Hood 讲道理:这种题我是绝对不去(敢)碰的.比赛时被这个题坑了一把,对于我这种不A不罢休的人来说就算看题解也要得到一个Accepted. 这题网上有很多题解,我自己是很难做出来的,于是参考了一下思路,确实很niub的一个题.先记录下来将来回来也有个参考. 题意:有n个人,每个人有一定数量的硬币a[i],每天将最富有的一个人的一枚硬币给最穷的那个人.求k天后最富有的人与最穷的人的差为多少. 思路:两遍二分啊,前所未见.因为k是固定的,我们先将所有人所能达到的上下界求出来,比如…

题目链接: http://codeforces.com/contest/672/problem/D 题意: 给你一个数组,每次操作,最大数减一,最小数加一,如果最大数减一之后比最小数加一之后要小,则取消操作,现在给你操作的次数,问操作之后最大数减最小数的最小值. 题解: 问题要求得是min(k次操作之后的最大数-k次操作之后的最小数),而这两个数可以独立求出来,我们先用二分求k次操作之后的最小数的最大取值,然后,再用二分求k次操作之后的最大数的最小可能取值. 代码: #include<algor…

题目链接:http://codeforces.com/contest/672/problem/D 有n个人,k个操作,每个人有a[i]个物品,每次操作把最富的人那里拿一个物品给最穷的人,问你最后贫富差距有多少. 先sort一下,最富的人很明显不会低于sum(a1 ~ an) / n , 所以二分一下最富人的最小值 看是否满足操作k. 最穷的人不会高于sum(a1 ~ an) / n + 1 ,所以二分一下最穷人的最大值 看是否满足操作k. (注意一点的是二分以后最穷的人的最大值和最富的人的最小值…

D. Robin Hood We all know the impressive story of Robin Hood. Robin Hood uses his archery skills and his wits to steal the money from rich, and return it to the poor. There are n citizens in Kekoland, each person has ci coins. Each day, Robin Hood wi…

Problems     # Name     A Summer Camp standard input/output 1 s, 256 MB    x3197 B Different is Good standard input/output 2 s, 256 MB    x2870 C Recycling Bottles standard input/output 2 s, 256 MB    x664 D Robin Hood standard input/output 1 s, 256…

模拟 A - Summer Camp #include <bits/stdc++.h> int a[1100]; int b[100]; int len; void init() { int i = 1, tot = 1; while (tot <= 1010) { int t = i, c = 0; while (t) { b[c++] = t % 10; t /= 10; } for (int i=c-1; i>=0; --i) { a[tot++] = b[i]; } i++…

672A Summer Camp 题意: 1-n数字连成一个字符串, 给定n , 输出字符串的第n个字符.n 很小, 可以直接暴力. Code: #include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; int begin = 1; for(;;) { int dig = log10(begin) + 1; if(n <= dig) { char ss[10]; sprintf(ss…