HDU 4726 Kia's Calculation

Kia's Calculation

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

Total Submission(s): 615    Accepted Submission(s): 174

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

159583036

 

Sample Output

Case #1: 8984

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup2

分析：贪心。

Code：

#include <algorithm>#include <iostream>#include <cstring>#include <cstdio>#define Min(a,b) ((a)<(b)?(a):(b))using namespace std;const int maxn=1000005;int s[2][10],ans[maxn];char s1[maxn],s2[maxn];int T,cnt,cas=1;int get_x(int x) {    int num=0,mx;    for(int i=0;i<=9;i++) {        for(int j=0;j<=9;j++) {            int tmp=(i+j)%10;            if(tmp==x) {                mx=Min(s[0][i],s[1][j]);                num+=mx;                s[0][i]-=mx;                s[1][j]-=mx;                mx=Min(s[1][i],s[0][j]);                num+=mx;                s[1][i]-=mx;                s[0][j]-=mx;            }        }    }    return num;}bool get_head() {    for(int k=9;k>0;k--) {        for(int i=1;i<=9;i++) {            for(int j=1;j<=9;j++) {                int tmp=(i+j)%10;                if(tmp==k) {                    if(s[0][i] && s[1][j]) {                        s[0][i]--; s[1][j]--;                        ans[0]=k;                        return true;                    }                    else if(s[0][j] && s[1][i]) {                        s[0][j]--; s[1][i]--;                        ans[0]=k;                        return true;                    }                }            }        }    }    return false;}void solve(){    for(int i=9;i>0;i--) {        int num=get_x(i);        while(num--) {            ans[cnt++]=i;        }    }}int main(){    scanf("%d",&T);    while(T--) {        scanf("%s %s",s1,s2);        int len=strlen(s1);        memset(s,0,sizeof(s));        for(int i=0;i<len;i++) {            s[0][s1[i]-'0']++;            s[1][s2[i]-'0']++;        }        cnt=1;        printf("Case #%d: ",cas++);        if(!get_head()) {            printf("0\n");            continue;        }        solve();        for(int i=0;i<cnt;i++) printf("%d",ans[i]);        if(cnt<len){            while(cnt<len) {                putchar('0');                cnt++;            }        }        puts("");    }    return 0;}