codeforces 373 A - Efim and Strange Grade(算数模拟) 原题:Efim and Strange Grade 题意:给出一个n位的实型数,你可以选择t次在任意位进行四舍五入的进位,求最大结果. 解法:这道题一定不能忽略数位计算时本身带来的进位,如果我们要改变这个数的大小,一定是在最先的那个出现5以上的数字进行四舍五入,之后的t次允许我们多次四舍五入,如果自然进位则不消耗t. 最后一点,整数位的进位也是需要考虑的 #include <cstdio> #inc…
C. Efim and Strange Grade 题目连接: http://codeforces.com/contest/719/problem/C Description Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. First he got disappoin…
题目链接:http://codeforces.com/problemset/problem/719/C C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a…
Codeforces 718A Efim and Strange Grade 程序分析 jerry的程序 using namespace std; typedef long long ll; string buf; int i; void up(int at) { at--; if (at < 0) { buf = '1' + buf; i++; return; } if (buf[at] == '.') at--; buf[at]++; if (buf[at] == '9'+1) { buf[…
C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
C. Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any po…
Efim and Strange Grade time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any posit…
time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Efim just received his grade for the last test. He studies in a special school and his grade can be equal to any positive decimal fraction. F…
题意:给定一个浮点数,让你在时间 t 内,变成一个最大的数,操作只有把某个小数位进行四舍五入,每秒可进行一次. 析:贪心策略就是从小数点开始找第一个大于等于5的,然后进行四舍五入,完成后再看看是不是还可以,一循环下去,直到整数位,或者没时间了. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <…
题目链接:http://codeforces.com/contest/842/problem/D 题解:像这种求一段异或什么的都可以考虑用字典树而且mex显然可以利用贪心+01字典树,和线段树差不多就是比较节点总数和有的数字数比较有限向左边转移. 然后这个异或其实可以利用一个数num与一个一个的x异或然后求异或的mex也是容易的只要判断当前二进制位是1那么左右节点拥有的数字数互换(不用真的互换 只要用转移体现出来就行了).这里的01字典树写法是类似线段树且利用递归的方法. #include <i…