HDU 4726 Kia's Calculation热身赛2 1011题(贪心 每次都找最大位的放在前面)
Kia's Calculation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 268 Accepted Submission(s): 73
Problem Description
Doctor Ghee is teaching Kia how to calculate the sum of two integers. But Kia is so careless and alway forget to carry a number when the sum of two digits exceeds 9. For example, when she calculates 4567+5789, she will get 9246, and for 1234+9876, she will get 0. Ghee is angry about this, and makes a hard problem for her to solve:
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 ?
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 ?
Input
The rst line has a number T (T <= 25) , indicating the number of test cases.
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 10 6, and without leading zeros.
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 10 6, and without leading zeros.
Output
For test case X, output "Case #X: " first, then output the maximum possible sum without leading zeros.
Sample Input
1 5958 3036
Sample Output
Case #1: 8984
Source
2013 ACM/ICPC Asia Regional Online —— Warmup2
题目大意：
给你两个串，数位可以相互交换，问你两个串随意交换能组成最大和，注意这个加法是不进位的，而且0不能当头。
解题思路：贪心思想。每次都找两个和最大的放在最前面，第一位比较特殊都不可以取0，主要是先统计一下两个串0~9的个数。然后第一位确定了之后，如果第一位是0，就可以直接输出0了。每次在两个串里面找数位相加组成最大的，然后每次直接相当于批处理，至少让一个个数变成0。这样最多只会循环几十次，然后就break掉了。不过当时竟然跳不出循环。最后才发现是tmp取-1，因为到最后数位相加可能真的最大就是0.debug了半天。。。详见代码。
题目地址：Kia's Calculation
AC代码：
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<cstdio>
using namespace std;
int a[10];
int b[10];
char t[1000005];
char res[1000005];
void debug() //真心的，debug了很久
{
int i;
for(i=0;i<10;i++)
cout<<a[i]<<" ";
cout<<endl;
for(i=0;i<10;i++)
cout<<b[i]<<" ";
cout<<endl;
}
int main()
{
int tes,i,len,j,k;
scanf("%d",&tes);
int cas=0;
while(tes--)
{
for(i=0;i<10;i++)
{
a[i]=0,b[i]=0;
}
scanf("%s",t);
len=strlen(t);
for(i=0;i<len;i++) //统计第一串的0~9的个数
a[t[i]-'0']++;
scanf("%s",t);
for(i=0;i<len;i++) //统计第二串的0~9的个数
b[t[i]-'0']++;
int tmp=-1; //这个地方真是个大坑，开始写的0
//debug();
int m,n;
for(i=1;i<10;i++) //先找第一位，第一位不能是0，所以都是从1到9找的
{
if(a[i])
{
for(j=1;j<10;j++)
{
if(b[j])
{
if((i+j)%10>tmp)
{
tmp=(i+j)%10;
m=i,n=j;
}
}
}
}
}
res[0]='0'+tmp; //第一位找到
if(res[0]=='0') //如果第一位是0，说明后面也只能是0，直接输出0就可以了
{
printf("Case #%d: 0\n",++cas);
continue;
}
a[m]--,b[n]--; //两个串相应的数字个数减少一个
int p=1;
while(1)
{
int m,n;
tmp=-1;
int cnt=0;
for(i=0;i<10;i++)
if(!a[i])
cnt++;
if(cnt==10)
break;
for(i=0;i<10;i++)
{
if(a[i])
{
for(j=9;j>=0;j--)
{
if(b[j])
{
int tt=(i+j)%10;
if(tt>tmp)
{
tmp=tt;
m=i,n=j;
}
}
}
}
}
int tm=a[m]<b[n]?a[m]:b[n];
a[m]-=tm,b[n]-=tm; //每次找到之后直接让一个个数或者两个个数都变为0
for(k=p;k<p+tm;k++)
res[k]='0'+tmp;
p+=tm;
}
//debug();
res[p]='\0';
printf("Case #%d: %s\n",++cas,res);
}
return 0;
}
/*
23
5958
3036
1234124141241241242142141245234242123123132144467874675850
4556143364476573566354365553645675435364754364635433264630
55
55
5
5
*/