hdu 5186 zhx's submissions

zhx's submissions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 648    Accepted Submission(s): 167

Problem Description

As one of the most powerful brushes, zhx submits a lot of code on many oj and most of them got AC.
One day, zhx wants to count how many submissions he made on n ojs. He knows that on the ith oj, he made ai submissions. And what you should do is to add them up.
To make the problem more complex, zhx gives you n B−base numbers and you should also return a B−base number to him.
What's more, zhx is so naive that he doesn't carry a number while adding. That means, his answer to 5+6 in 10−base is 1. And he also asked you to calculate in his way.

 

Input

Multiply test cases(less than 1000). Seek EOF as the end of the file.
For each test, there are two integers n and B separated by a space. (1≤n≤100, 2≤B≤36)
Then come n lines. In each line there is a B−base number(may contain leading zeros). The digits are from 0 to 9 then from a to z(lowercase). The length of a number will not execeed 200.

 

Output

For each test case, output a single line indicating the answer in B−base(no leading zero).

 

Sample Input

2 3221 42333 16abbccd

 

Sample Output

123314

这个题有 好几个比较坑的地方，

（1），，结果是0

（2），，根据我的算法，注意s2>s1（长度上）的情况

（3），，注意要给s1末尾加上\0

#include<iostream>#include<math.h>#include<fstream>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;char s1[220],s2[220];char s3[220];int b;void PRINT(char s[]){    bool flag=0;    for(unsigned int i=0;i<strlen(s);i++)    {        if(s[i]=='0'&&flag==0)  continue;        else        {            flag=1;            printf("%c",s[i]);        }    }    if(flag==0)  cout<<"0";    printf("\n");}void ADD(char s1[],char s2[]){    if(strlen(s1)>=strlen(s2))        for(int i=strlen(s1)-1,j=strlen(s2)-1;i>=0&&j>=0;i--,j--)        {            int t1=s1[i]>'9'?s1[i]-'a'+10:s1[i]-'0';            int t2=s2[j]>'9'?s2[j]-'a'+10:s2[j]-'0';            int t3=(t1+t2)%b;            if(t3>9)                s1[i]=t3-10+'a';            else                s1[i]=t3+'0';        }    else    {        for(int i=strlen(s1)-1,j=strlen(s2)-1;i>=0&&j>=0;i--,j--)        {            int t1=s1[i]>'9'?s1[i]-'a'+10:s1[i]-'0';            int t2=s2[j]>'9'?s2[j]-'a'+10:s2[j]-'0';            int t3=(t1+t2)%b;            if(t3>9)                s2[j]=t3-10+'a';            else                s2[j]=t3+'0';        }        for(int i=0;i<strlen(s2);i++)            s1[i]=s2[i];        s1[strlen(s2)]='\0';    }}int main(){    //freopen("1.txt","r",stdin);    int n;    while(scanf("%d %d",&n,&b)!=EOF)    {        if(n==1)        {            scanf("%s",s1);            PRINT(s1);        }        else        {            scanf("%s",s1);            scanf("%s",s2);            ADD(s1,s2);            for(int i=3;i<=n;i++)            {                scanf("%s",s2);                ADD(s1,s2);            }            PRINT(s1);        }    }    return 0;}