K - Kia's Calculation(贪心)
Kia's Calculation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Now Kia has two integers A and B, she can shuffle the digits in each number as she like, but leading zeros are not allowed. That is to say, for A = 11024, she can rearrange the number as 10124, or 41102, or many other, but 02411 is not allowed.
After she shuffles A and B, she will add them together, in her own way. And what will be the maximum possible sum of A "+" B ?
For each test case there are two lines. First line has the number A, and the second line has the number B.
Both A and B will have same number of digits, which is no larger than 106, and without leading zeros.
5958
3036
算法:贪心
题解:根据题目意思来,我们只要将所有的数字出现的个数都记录一下(桶排),然后你就遍历依次相加取最大就行了。要注意的你需要单独判断第一个数,它不能有0,因为你的变化是不能把0变成第一位的,然后你还要注意的是,你的结果不能有前导0。
#include <iostream>
#include <cstdio>
#include <memory.h> using namespace std; int visa[], visb[];
char ans[];
string a, b; int main() {
int T;
int cas = ;
scanf("%d", &T);
while(T--) {
for(int i = ; i < ; i++) {
visa[i] = visb[i] = ;
}
cin >> a >> b;
int lena = a.size();
int lenb = b.size();
for(int i = ; i < lena; i++) {
visa[a[i] - '']++;
}
for(int i = ; i < lenb; i++) {
visb[b[i] - '']++;
}
int posa, posb, maxx = -;
for(int i = ; i < ; i++) { //找出第一个数
for(int j = ; j < ; j++) {
if(i != && j != && visa[i] && visb[j] && maxx < (i + j) % ) {
maxx = (i + j) % ;
posa = i;
posb = j;
}
}
}
int len = ;
printf("Case #%d: ", ++cas);
if(maxx >= ) { //如果第一个数存在,则存储下来
ans[len++] = maxx + '';
visa[posa]--;
visb[posb]--;
}
for(int k = ; k >= ; k--) { //寻找之后的数字,每次取最大
for(int i = ; i < ; i++) {
for(int j = ; j < ; j++) {
while(visa[i] > && visb[j] > && (i + j) % == k) {
visa[i]--;
visb[j]--;
ans[len++] = k + '';
}
}
}
}
int mark = ;
for(int i = ; i < len; i++) { //需要判断前导0
if(mark && i == len - ) {
printf("%c", ans[i]);
} else if(mark && ans[i] != '') {
mark = ;
printf("%c", ans[i]);
} else if(!mark) {
printf("%c", ans[i]);
}
}
printf("\n");
}
return ;
}
K - Kia's Calculation(贪心)的更多相关文章
- K - Kia's Calculation (贪心)
Kia's Calculation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- HDU-4726 Kia's Calculation 贪心
题目链接:http://acm.hdu.edu.cn/userstatus.php?user=zhsl 题意:给两个大数,他们之间的加法法则每位相加不进位.现在可以对两个大数的每位重新排序,但是首位不 ...
- 贪心 HDOJ 4726 Kia's Calculation
题目传送门 /* 这题交给队友做,做了一个多小时,全排列,RE数组越界,赛后发现读题读错了,囧! 贪心:先确定最高位的数字,然后用贪心的方法,越高位数字越大 注意:1. Both A and B wi ...
- HDU 4726 Kia's Calculation (贪心算法)
Kia's Calculation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) T ...
- HDU 4726 Kia's Calculation(贪心)
Kia's Calculation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- Kia's Calculation hdu4726
Kia's Calculation Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others ...
- ACM学习历程—HDU 4726 Kia's Calculation( 贪心&&计数排序)
DescriptionDoctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so carel ...
- HDU 4726 Kia's Calculation(贪心构造)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4726 题意:给出两个n位的数字,均无前缀0.重新排列两个数字中的各个数,重新排列后也无前缀0.得到的两 ...
- Kia's Calculation(HDU 4267)
Problem Description Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is ...
随机推荐
- MySQL 聚合函数与count()函数
一.MySQL中的聚合函数 MySQL 5.7文档的章节:12.20.1 Aggregate (GROUP BY) Function “聚合/组合”函数(group (aggregate) funct ...
- 怎样创建一个canvas画布环境
1. 由于canvas画布在网页中, 所以需要在html中添加canvas标签: <!DOCTYPE html> <html lang="en"> < ...
- PLSQL PL/SQL Developer Oracle 使用技巧 常用设置 卡顿问题 病毒防范( 附带:配置文件)
相关工具版本: PL/SQL Developer: 9.0.4.1644 Oracle : Oracle Database 10g Enterprise Edition Release 10.2.0. ...
- Linux学习(一)-安装vm虚拟机以及如何在虚拟机上安装Centos系统
(一)基本说明 学习Linux需要一个环境,我们需要创建一个虚拟机,然后在虚拟机上安装一个Centos系统来学习. 1)安装软件vm12; 2)通过vm12创建一个虚拟机空间; 3)在vm12创建好的 ...
- 【jekins】jenkins构建触发
一.定时构建的语法 * * * * *(五颗星,中间用空格隔开)第一颗*表示分钟,取值0~59第二颗*表示小时,取值0~23第三颗*表示一个月的第几天,取值1~31第四颗*表示第几月,取值1~12第五 ...
- printf固定一行打印倒计时的实现
@2019-07-15 [小记] #include<stdlib.h> #include <stdio.h> #include <time.h> #include ...
- sql 拼接字符串单条拆分多条
SELECT * FROM ( SELECT A.WS_ID , B.NEXT_OPERATOR FROM ( SELECT WS_ID , [NEXT_OPERATOR] = CONVERT(XML ...
- Repeater POJ - 3768 (分形)
Repeater POJ - 3768 Harmony is indispensible in our daily life and no one can live without it----may ...
- 【Java基础 项目实例--Bank项目5】Account 和 customer 对象等 继承、多态、方法的重写
延续 Java基础 项目实例--Bank项目4 实验要求 实验题目 5: 在银行项目中创建 Account 的两个子类:SavingAccount 和 CheckingAccount 实验目的: 继承 ...
- [转]makefile学习
原文: http://blog.fatedier.com/2014/09/08/learn-to-write-makefile-01/ -------------------------------- ...