hdu 4726(贪心)

时间:2022-01-11 22:42:20

Kia's Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3291    Accepted Submission(s): 703

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 ?
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 106, 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
题意:给出两个长度不超过 10^6 的数字串,数字串可以打乱后随机组合,但是打乱重组后不能有前导 0,将这两个数字串相加,相加的规则是每一位相加,不实现进位.问能够得到最大的结果串是多少?
题解:贪心求解,从高位到低位,每次贪心选择相加起来最大的数字,最高位要单独处理,因为最高位要求被加数,加数,结果都不能为 0.然后在贪心的过程中不能去枚举加数和被加数,只能枚举结果,不然会超时. 当位数只有1,加数或者被加数中间有一个为0时单独处理.
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
const int N = ;
char str1[N],str2[N];
int num1[],num2[];
int res[N];
int main()
{
int tcase,t = ;
scanf("%d",&tcase);
while(tcase--)
{ scanf("%s%s",str1,str2);
if(strcmp(str1,"")==){
printf("Case #%d: ",t++);
printf("%s\n",str2);
continue;
}
if(strcmp(str2,"")==){
printf("Case #%d: ",t++);
printf("%s\n",str1);
continue;
}
int len = strlen(str1);
memset(num1,,sizeof(num1));
memset(num2,,sizeof(num2));
for(int i=; i<len; i++)
{
num1[str1[i]-'']++;
num2[str2[i]-'']++;
}
int high = -,x,y;
for(int i=; i<=; i++)
{
for(int j=; j<=; j++)
{
if(num1[i]&&num2[j]&&high<(i+j)%)
{
x = i;
y = j;
high = (i+j)%;
}
}
}
num1[x]--;
num2[y]--;
int cnt = ,zero = ;
res[cnt++] = high;
if(high==) zero++;
printf("Case #%d: ",t++);
if(zero){
printf("0\n");
continue;
}
for(int l=; l<len; l++)
{
/*
TLE
for(int i=0; i<=9; i++)
{
for(int j=0; j<=9; j++)
{
if(num1[i]&&num2[j]&&MAX<(i+j)%10)
{
x = i;
y = j;
MAX = (i+j)%10;
}
}
}*/
bool flag = true;
for(int i=;i>=&&flag;i--){
for(int j=;j<=&&flag;j++){
if(i-j<&&num1[j]&&num2[i-j+]){
num1[j]--;
num2[i-j+]--;
res[cnt++] = i;
flag = false;
}else if(i-j>=&&num1[j]&&num2[i-j]){
num1[j]--;
num2[i-j]--;
res[cnt++] = i;
flag = false;
}
}
} }
for(int i=;i<cnt;i++){
printf("%d",res[i]);
}
printf("\n");
}
return ;
}