HDU - 5186 - zhx&#39;s submissions （精密塔尔苏斯）

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## HDU - 5186 - zhx&#39;s submissions （精密塔尔苏斯）

eddie小英俊 2017-11-15 21:18:00 浏览452

# 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 Bbase numbers and you should also return a Bbase 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 10base 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. (1n1002B36)
Then come n lines. In each line there is a Bbase 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 Bbase(no leading zero).

Sample Input

2 3 2 2 1 4 233 3 16 ab bc cd

Sample Output

1 233 14

Source

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);
}
print(ans);
}
return 0;
}

#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);
}
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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