1001
首先读进来的时候把字母和数字都转换成0到35的数字,加起来直接取模,算出答案。 坑点是只有1个数的情况,还有答案等于0的时候也要输出一行一个0。
注意去掉前导0,因为求和过程也有可能产生0,所以求完和在去0。
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#include <set>
using namespace std;
#define read() freopen("data.in", "r", stdin)
#define write() freopen("data.out", "w", stdout)
#define rep( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i )
#define For( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i )
#define clr( a , x ) memset ( a , x , sizeof a )
#define cpy( a , x ) memcpy ( a , x , sizeof a )
#define _max(a,b) ((a>b)?(a):(b))
#define _min(a,b) ((a<b)?(a):(b))
#define LL long long
const int maxNumber=205;
int sum[maxNumber];
char s[maxNumber];
int main()
{
//read();
int n,b;
while(cin>>n>>b)
{
int temp = 0;
clr(sum,0);
while(n--)
{
scanf("%s",s);
int len = strlen(s);
reverse(s,s+len);
for (int i = 0; i < len; ++i)
{
if (s[i]>=‘a‘&&s[i]<=‘z‘)
{
sum[i]+=(s[i]-‘a‘+10);
}else
{
sum[i]+=s[i]-‘0‘;
}
}
temp = _max(temp,len);
}
for (int i = 0; i < temp; ++i)
{
sum[i] %= b;
}
while (temp > 1 && sum[temp-1]==0)
{
--temp;
}
for (int i = temp-1; i >= 0; --i)
{
if (sum[i] < 10)
{
printf("%d",sum[i] );
}else
{
printf("%c",sum[i]-10+‘a‘ );
}
}
printf("\n");
}
return 0;
}
原文:http://www.cnblogs.com/acmsummer/p/4363543.html