zhx's submissions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 540 Accepted Submission(s): 146
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.
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.
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).
leading zero).
Sample Input
2 3
2
2
1 4
233
3 16
ab
bc
cd
Sample Output
1
233
14
Source
思路：就是不進位的大數相加啦，要注意當結果為0時輸出一個0。之前我還做過一個差點兒相同的，上次注意到了，。這次竟然沒注意到o(╯□╰)o.........
疑問：為何執行時間900多ms，並且還可能會T，把cstdio改為stdio.h時間就降下來了。直接變為100多ms，害的我還檢查半天。。。可是這是為什麼？？？？？？
搞了半天我發現使用g++環境提交的沒過。而用c++環境就過啦（以後再HDU做題還是用c++環境吧。醉啦）
據說g++用scanf由於輸入太慢而要開掛（難道和cin減速一個性質？？）。。，。貌似是這種，以後再試試
void gn(int &x){
char c;while((c=getchar())<'0'||c>'9');x=c-'0';
while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';
}
AC程式碼①（100+ms。g++環境）：
#include <stdio.h>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; char ans[205];
char t[205]; void fun(char ans[], char t[]) {
int len = strlen(t);
for(int i = 0; i < len; i++) {
ans[i] = t[len - 1 - i];
}
} void swap(char t[]) {
int len = strlen(t);
for(int i = 0; i < len / 2; i++) {
char m = t[i];
t[i] = t[len - 1 - i];
t[len - 1 - i] = m;
}
} void add(char ans[], char t[], int B) {
int t1, t2, t3;
int len = strlen(t);
for(int i = 0; i < len; i++) {
if(ans[i] <= 'z' && ans[i] >= 'a') t1 = (int)(ans[i] - 'a' + 10);
else t1 = ans[i] - '0';
if(t[i] <= 'z' && t[i] >= 'a') t2 = (int)(t[i] - 'a' + 10);
else t2 = t[i] - '0';
t3 = (t1 + t2) % B;
if(t3 >= 10) ans[i] = (char)(t3 - 10 + 'a');
else ans[i] = (char)(t3 + '0');
}
} void print(char ans[]) {
int flag = 0, p;
for(int i = 204; i >= 0; i--) {
if(ans[i] != '0') {
printf("%c", ans[i]);
flag = 1;
}
else if(ans[i] == '0' && flag) printf("0");
}
if(flag == 0) printf("0");
printf("\n");
} int main() {
int n, B;
while(scanf("%d %d", &n, &B) != EOF) {
for(int i = 0; i< 205; i++) ans[i] = '0'; scanf("%s", t);
fun(ans, t);
for(int i = 0; i < n-1; i++) {
scanf("%s", t);
swap(t);
add(ans, t, B);
}
print(ans);
}
return 0;
}
程式碼②（900+ms or TLE。g++環境）：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std; char ans[205];
char t[205]; void fun(char ans[], char t[]) {
int len = strlen(t);
for(int i = 0; i < len; i++) {
ans[i] = t[len - 1 - i];
}
} void swap(char t[]) {
int len = strlen(t);
for(int i = 0; i < len / 2; i++) {
char m = t[i];
t[i] = t[len - 1 - i];
t[len - 1 - i] = m;
}
} void add(char ans[], char t[], int B) {
int t1, t2, t3;
int len = strlen(t);
for(int i = 0; i < len; i++) {
if(ans[i] <= 'z' && ans[i] >= 'a') t1 = (int)(ans[i] - 'a' + 10);
else t1 = ans[i] - '0';
if(t[i] <= 'z' && t[i] >= 'a') t2 = (int)(t[i] - 'a' + 10);
else t2 = t[i] - '0';
t3 = (t1 + t2) % B;
if(t3 >= 10) ans[i] = (char)(t3 - 10 + 'a');
else ans[i] = (char)(t3 + '0');
}
} void print(char ans[]) {
int flag = 0, p;
for(int i = 204; i >= 0; i--) {
if(ans[i] != '0') {
printf("%c", ans[i]);
flag = 1;
}
else if(ans[i] == '0' && flag) printf("0");
}
if(flag == 0) printf("0");
printf("\n");
} int main() {
int n, B;
while(scanf("%d %d", &n, &B) != EOF) {
for(int i = 0; i< 205; i++) ans[i] = '0'; scanf("%s", t);
fun(ans, t);
for(int i = 0; i < n-1; i++) {
scanf("%s", t);
swap(t);
add(ans, t, B);
}
print(ans);
}
return 0;
}
AC程式碼③：
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; #define maxn 205
char tmp[maxn][maxn], ans[maxn][maxn], ch[50];
int to[maxn]; void init() {
memset(ch, 0, sizeof(ch));
memset(to, 0, sizeof(to));
for(int i = 0; i <= 35; i++) {
if(i <= 9) ch[i] = i + '0', to[i + '0'] = i;
else ch[i] = i - 10 + 'a', to[i - 10 + 'a'] = i;
}
} int main() {
int n, B;
init();
while(~scanf("%d %d", &n, &B)) {
memset(ans, 0, sizeof(ans));
memset(tmp, 0, sizeof(tmp)); for(int i = 1; i <= n; i++) {
scanf("%s", tmp[i]);
int len = strlen(tmp[i]);
for(int j = 0; j < len; j++) {
ans[i][j] = tmp[i][len-1-j];
}
} int flag = 0;
for(int i = maxn - 1; i >= 0; i--) {
int t = 0;
for(int j = 1; j <= n; j++) {
t += to[ans[j][i]];
}
t %= B;
if(t) flag = 1;
if(flag) printf("%c", ch[t]);
}
if(!flag) printf("0");
printf("\n");
}
return 0;
}
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