码迷,mamicode.com
首页 > 其他好文 > 详细

K - Kia's Calculation(贪心)

时间:2019-08-22 11:02:55      阅读:78      评论:0      收藏:0      [点我收藏+]

标签:注意   结果   第一个   dig   shuf   amp   cin   意思   ros   

Kia‘s Calculation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)


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

 

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[15], visb[15];
char ans[1000007];
string a, b;

int main() {
    int T;
    int cas = 0;
    scanf("%d", &T);
    while(T--) {
        for(int i = 0; i < 10; i++) {
            visa[i] = visb[i] = 0;
        }
        cin >> a >> b;
        int lena = a.size();
        int lenb = b.size();
        for(int i = 0; i < lena; i++) {
            visa[a[i] - 0]++;
        }
        for(int i = 0; i < lenb; i++) {
            visb[b[i] - 0]++;
        }
        int posa, posb, maxx = -1;
        for(int i = 0; i < 10; i++) {    //找出第一个数
            for(int j = 0; j < 10; j++) {
                if(i != 0 && j != 0 && visa[i] && visb[j] && maxx < (i + j) % 10) {
                    maxx = (i + j) % 10;
                    posa = i;
                    posb = j;
                }
            }
        }
        int len = 0;
        printf("Case #%d: ", ++cas);
        if(maxx >= 0) {    //如果第一个数存在,则存储下来
            ans[len++] = maxx + 0;
            visa[posa]--;
            visb[posb]--;
        }
        for(int k = 9; k >= 0; k--) {        //寻找之后的数字,每次取最大
            for(int i = 0; i < 10; i++) {
                for(int j = 0; j < 10; j++) {
                    while(visa[i] > 0 && visb[j] > 0 && (i + j) % 10 == k) {
                        visa[i]--;
                        visb[j]--;
                        ans[len++] = k + 0;
                    }
                }
            }
        }
        int mark = 1;
        for(int i = 0; i < len; i++) {    //需要判断前导0
            if(mark && i == len - 1) {
                printf("%c", ans[i]);
            } else if(mark && ans[i] != 0) {
                mark = 0;
                printf("%c", ans[i]);
            } else if(!mark) {
                printf("%c", ans[i]);
            }
        }
        printf("\n");
    }
    return 0;
}

 

K - Kia's Calculation(贪心)

标签:注意   结果   第一个   dig   shuf   amp   cin   意思   ros   

原文地址:https://www.cnblogs.com/buhuiflydepig/p/11392832.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!