https://vjudge.net/problem/Gym-102058M
Mr. Jeong really loves coke. He loves so much that he drinks coke everyday without exception. One day, he decided to open a coke contest in Daejeon. To the winner, a box of cokes will be given!
N people participate in the contest. Each participant is given K mL of coke, and the one who finishes his/her coke earliest wins the contest. But it is painful to drink the whole coke in one time, so each person repeats drinking and taking a rest. More specifically, as soon as the contest starts, the ith participant starts to drink for ti seconds, then takes a rest for si seconds, and repeats this process until no more coke remains. Moreover, everyone drinks A mL per second. The contest is over if one of the participants finished his/her coke.
Given the infomation of N participants, determine after how many seconds the contest is finished, since the contest begins.
Input
The input starts with the first line containing three integers N (2 ≤ N ≤ 1000), K (1 ≤ K ≤ 10000), and A (1 ≤ A ≤ 100). The ith of the next N lines contains two integers ti (1 ≤ ti ≤ 100) and si (1 ≤ si ≤ 100), the information of ith participant. K is a multiple of A.
Output
Write a single integer, the answer to the question.
Examples
Input
2 100 1 10 5 5 10
Output
145
Input
4 100 2 30 30 49 2 50 50 20 10
Output
50
题意分析：
有一个喝汽水比赛，每个人喝 t 秒后必须休息 s 秒， 一共K毫升汽水，每秒可以喝 a 毫升，一旦有一个人喝完汽水，比赛结束，求比赛结束的时间。
解题思路：
对每个人喝完时间进行比较，找出最短的那一个， 如果一个人喝完汽水，那后面的等待时间就不用有。
第一次 sum=(s+t)*(k/t); 后面的 k/t 没加括号错了1次。
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
int n, k, a, ans, sum, t, s;
while(scanf("%d%d%d", &n ,&k, &a)!=EOF)
{
ans=99999999;
k/=a;
while(n--)
{
scanf("%d%d", &t, &s);
sum=(s+t)*(k/t);
if(k%t==0)
sum-=s;
else
sum+=k%t;
ans=min(ans, sum);
}
printf("%d
", ans);
}
return 0;
}