HDU 5163 Taking Bus
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Taking Bus
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 418 Accepted Submission(s): 169
Problem Description
Bestland has a very long road and there are n bus station along the road, which are numbered 1 to n from left to right. There are m persons wanting to take the bus to some other station. You task is to find the time needed for each person. Note: All the other information you need is below. Please read the statment carefully.
Input
There are multiple test cases. The first line of input contains an integer T (1≤T≤60) , indicating the number of test cases. For each test case: The first line contains two integers n and m (2≤n,m≤105) , indicating the number of bus stations and number of people. In the next line, there are n?1 integers, d1,d2,…,dn?1 (1≤di≤109 ). The i -th integer means the distance between bus station i and i+1 is di (1≤i<n ). In the next m lines, each contains two integers xi and yi (1≤xi,yi≤n,xi≠yi ), which means i -th person is in bus station xi and wants goto bus station yi . (1≤i≤m)
What else you should know is that for thei -th person, the bus starts at bus station ((i?1) mod n)+1 and drives to right. When the bus arrives at station n , it will turn around and drive from right to left. Similarly, When the bus arrives at station 1 , it will turn around and drive from left to right. You can assume that the bus drives one meter per second. And you should only consider the time that the bus drives and ignore the others.
What else you should know is that for the
Output
For each person, you should output one integer which is the minimum time needed before arriving bus station yi .
Sample Input
17 32 3 4 3 4 51 74 55 4
Sample Output
211028HintFor the first person, the bus starts at bus station 1, and the person takes in bus at time 0. After 21 seconds, the bus arrives at bus station 7. So the time needed is 21 seconds. For the second person, the bus starts at bus station 2. After 7 seconds, the bus arrives at bus station 4 and the person takes in the bus. After 3 seconds, the bus arrives at bus station 5. So the time needed is 10 seconds. For the third person, the bus starts at bus station 3. After 7 seconds, the bus arrives at bus station 5 and the person takes in the bus. After 9 seconds, the bus arrives at bus station 7 and the bus turns around. After 12 seconds, the bus arrives at bus station 4. So the time needed is 28 seconds.
Source
BestCoder Round #27
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hujie | We have carefully selected several similar problems for you: 5165 5164 5161 5160 5159
维护前缀和的思想 代码如下。
#include<iostream>
#include<stdio.h>
#define ll long long
using namespace std;
ll ans;
int n,m;
ll a;
ll s[100005];
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int T,t;
scanf("%d",&T);
while(T--)
{
int x,y;
scanf("%d %d",&n,&m);
s[1]=0;
for(int i=1;i<n;i++)
{
scanf("%I64d",&a);
s[i+1]=s[i]+a;
}
for(int i=1;i<=m;i++)
{
ans=0;
scanf("%d %d",&x,&y);
t=(i-1)%n+1;
if(x<=y)
{
if(t<=x)
{
ans=s[y]-s[t];
}
else
{
ans=s[n]+s[y]-s[t]+s[n];
}
}
else
{
ans=s[n]-s[t]+s[n]-s[y];
}
printf("%I64d\n",ans);
}
}
return 0;
}
好坑啊。变量名写重复了,交了n多遍都是TLE ...要不还能长点分。
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