BestCoder Round #85题解报告

1.sum(杭电5776)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 772    Accepted Submission(s): 363

Problem Description

Given a sequence, you're asked whether there exists a consecutive subsequence whose sum is divisible by m. output YES, otherwise output NO

 

Input

The first line of the input has an integer T ( 1≤T≤10 ), which represents the number of test cases.
For each test case, there are two lines:
1.The first line contains two positive integers n, m ( 1≤n≤100000 , 1≤m≤5000 ).
2.The second line contains n positive integers x ( 1≤x≤100 ) according to the sequence.

 

Output

Output T lines, each line print a YES or NO.

 

Sample Input

  

   2
3 3
1 2 3
5 7
6 6 6 6 6
  

 

Sample Output

  

   YES
NO
  

 

Source

BestCoder Round #85

 

【题意】
给定一个数列,问是否存在连续子串的和为m的倍数

【类型】
预处理前缀和

【分析】
所谓预处理前缀和

顾名思义,就是将前i项的和相加对m取模保存至sum[i]

这时,当某两个位置的模数一样时,此区间内所有数之和必定为m的倍数

为什么呢?证明一下

假设sum[i]=sum[j],则

特别的,若某前缀和取模后为0,意味着数列从头到当前位置所有元素之和已为m的倍数

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wange2014   |   We have carefully selected several similar problems for you:   5780  5779  5778  5777  5775 

#include<stdio.h>
#include<string.h>
int main()
{
int t,n,m,i;
int a[100010],b[100010],visit[100010];
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
b[i]=sum%m;
}
memset(visit,0,sizeof(visit));
for(i=0;i<n;i++)
{
if(b[i]==0||visit[b[i]])
{
printf("YES\n");
break;
}
else
visit[b[i]]=1;
}
if(i==n)
printf("NO\n");
}
return 0;
}

 

2.domino(杭电5777)

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 539    Accepted Submission(s): 277

Problem Description

Little White plays a game.There are n pieces of dominoes on the table in a row. He can choose a domino which hasn't fall down for at most k times, let it fall to the left or right. When a domino is toppled, it will knock down the erect domino. On the assumption that all of the tiles are fallen in the end, he can set the height of all dominoes, but he wants to minimize the sum of all dominoes height. The height of every domino is an integer and at least 1.

 

Input

The first line of input is an integer T ( 1≤T≤10 )
There are two lines of each test case.
The first line has two integer n and k, respectively domino number and the number of opportunities.( 2≤k,n≤100000 )
The second line has n - 1 integers, the distance of adjacent domino d, 1≤d≤100000

 

Output

For each testcase, output of a line, the smallest sum of all dominoes height

 

Sample Input

   

    1
4 2
2 3 4
   

 

Sample Output

   

    9
   

 

Source

BestCoder Round #85

 

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wange2014

解题思路:

【题意】
n张多米诺骨牌排成一列,你可以选k张推倒,现给你相邻骨牌之间的距离,问在保证所有骨牌都能推倒的情况下,所有骨牌的高度之和最小为多少

【类型】
排序

【分析】
首先,在只推一次的情况下,骨牌i的高度hi必须满足hi>=di+1(di为骨牌i和骨牌i+1之间的距离)

而每多推一次,其实只能消去一个间距罢了

那为了骨牌高度之和尽可能小,那必定是要消去间距大的

故将间距排序之后,去掉最大的k-1个即可

特别的,当k>=n时,因为至少每个骨牌都可以单独推,所有骨牌高度与间距无关,只需满足题目要求的骨牌高度至少为1即可

需注意,结果可能会爆int