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…
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…
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…
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 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…
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[…
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 <…
剑指Offer:链表中倒数第k个结点[22] 题目描述 输入一个链表,输出该链表中倒数第k个结点. 解题思考 我们定义两个指针L和R,R事先移动K-1个位置,然后两者同时往后移动直到遇到R的下个节点为空,此时L节点的位置就是倒数第K个节点. Java题解 package linklist; public class FindKthToFail { public static ListNode FindKthToTail(ListNode head,int k) { if(head==null||…