BestCoder Round #33

A.

 zhx's submissions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 858    Accepted Submission(s): 233

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.

 

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.

 

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

BestCoder Round #33

题意：n个B进制的数相加。B进制输出结果。

解析：開始理解错题意了。最后才发现是每位相加时不产生进位。。。直接模拟就可以。

AC代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, m;

char s[1002][202];
char ans[202];
int len[202];

int get(char c){
	return (c >= '0' && c <= '9') ? c - '0' : c  - 'a' + 10;
}

char put(int x){
	return x < 10 ?
 x + '0' : x - 10 + 'a';
}

int main(){
//	freopen("in.txt", "r", stdin);
	while(scanf("%d%d", &n, &m)==2){
		int maxlen = 0;
		for(int i=0; i<n; i++){
			scanf("%s", s[i]);
			len[i] = strlen(s[i]);
			maxlen = max(maxlen, len[i]);
		}

		for(int i=0; i<maxlen; i++) ans[i] = '0';
		ans[maxlen] = 0;
		for(int j=1; j<=maxlen; j++){
			for(int i=0; i<n; i++){
				if(len[i] >= j){
					int cnt = maxlen - j;
					ans[cnt] = put((get(ans[cnt]) + get(s[i][len[i] - j])) % m);
				}
			}
		}
		int i = 0;
		while(ans[i] == '0' && i < maxlen - 1) i ++;
		printf("%s
", ans + i);
	}
	return 0;
}

PS：模拟到吐血。。

。

B.

zhx's contest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 833    Accepted Submission(s): 274

Problem Description

As one of the most powerful brushes, zhx is required to give his juniors n problems.
zhx thinks the ith problem's difficulty is i. He wants to arrange these problems in a beautiful way.
zhx defines a sequence {ai} beautiful if there is an i that matches two rules below:
1: a1..ai are monotone decreasing or monotone increasing.
2: ai..an are monotone decreasing or monotone increasing.
He wants you to tell him that how many permutations of problems are there if the sequence of the problems' difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him the answer module p.

 

Input

Multiply test cases(less than 1000). Seek EOF as the end of the file.
For each case, there are two integers n and p separated by a space in a line. (1≤n,p≤1018)

 

Output

For each test case, output a single line indicating the answer.

 

Sample Input

2 233
3 5

 

Sample Output

2
1

Hint
In the first case, both sequence {1, 2} and {2, 1} are legal.
In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1
 

 

Source

BestCoder Round #33

题意：给一个n，统计1~n的数构成的先递增后递减或先递减后递增的序列的个数。

解析：这样的情况仅仅须要考虑最大值的位置就可以，共同拥有2^(n-1)  * 2种，在两端时有单增或单减2种。可是前面考虑的时候反复了两次，故终于结果是2^n - 2.注意要特判n == 1时，直接输出结果就可以。

AC代码：

#include <cstdio>

#define LL __int64

LL q_mul(LL x, LL n, LL p){          //快乘，非常坑的是两数都不能直接相乘。据说会爆掉
    LL res = 0;
	while(n){
        if(n & 1) res = (res + x) % p;
        x = (x + x) % p;
        n >>= 1;
	}
	return res;
}

LL pow_mod(LL x, LL n, LL p){        //高速幂
	LL res = 1;
	while(n){
        if(n & 1) res = q_mul(res, x, p);   //计算res * x
        x = q_mul(x, x, p);                 //计算x * x
        n >>= 1;
	}
	return res;
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif //sxk

	LL n, p;
	while(scanf("%I64d%I64d", &n, &p)!=EOF){
		if(n == 1){
            printf("%I64d
", n % p);
            continue;
        }
		else printf("%I64d
", (pow_mod(2, n, p) - 2 + p) % p);
	}
	return 0;
}

未完。

。。待续