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