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