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.
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 3 2 2 1 4 233 3 16 ab bc cd
Sample Output
1 233 14
Source
思路：就是不进位的大数相加啦，要注意当结果为0时输出一个0。之前我还做过一个差点儿相同的，上次注意到了，。这次竟然没注意到o(╯□╰)o.........
疑问：为何执行时间900多ms，并且还可能会T，把cstdio改为stdio.h时间就降下来了。直接变为100多ms，害的我还检查半天。。。可是这是为什么？？？？？？
搞了半天我发现使用g++环境提交的没过。而用c++环境就过啦（以后再HDU做题还是用c++环境吧。醉啦）
据说g++用scanf由于输入太慢而要开挂（难道和cin减速一个性质？？）。。，。貌似是这种，以后再试试
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代码①（100+ms。g++环境）：
#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;
}
代码②（900+ms or TLE。g++环境）：
#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代码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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