Alice's Print Service
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2728 Accepted Submission(s): 708
Problem Description
Alice is providing print service, while the pricing doesn't seem to be reasonable, so people using her print service found some tricks to save money.
For example, the price when printing less than 100 pages is 20 cents per page, but when printing not less than 100 pages, you just need to pay only 10 cents per page. It's easy to figure out that if you want to print 99 pages, the best choice is to print an extra blank page so that the money you need to pay is 100 × 10 cents instead of 99 × 20 cents.
Now given the description of pricing strategy and some queries, your task is to figure out the best ways to complete those queries in order to save money.
For example, the price when printing less than 100 pages is 20 cents per page, but when printing not less than 100 pages, you just need to pay only 10 cents per page. It's easy to figure out that if you want to print 99 pages, the best choice is to print an extra blank page so that the money you need to pay is 100 × 10 cents instead of 99 × 20 cents.
Now given the description of pricing strategy and some queries, your task is to figure out the best ways to complete those queries in order to save money.
Input
The first line contains an integer T (≈ 10) which is the number of test cases. Then T cases follow.
Each case contains 3 lines. The first line contains two integers n, m (0 < n, m ≤ 105 ). The second line contains 2n integers s1, p1 , s2, p2 , ..., sn, pn (0=s1 < s2 < ... < sn≤ 109 , 109 ≥ p1 ≥ p2 ≥ ... ≥ pn ≥ 0).. The price when printing no less than si but less than si+1 pages is pi cents per page (for i=1..n-1). The price when printing no less than sn pages is pn cents per page. The third line containing m integers q1 .. qm (0 ≤ qi ≤ 109 ) are the queries.
Each case contains 3 lines. The first line contains two integers n, m (0 < n, m ≤ 105 ). The second line contains 2n integers s1, p1 , s2, p2 , ..., sn, pn (0=s1 < s2 < ... < sn≤ 109 , 109 ≥ p1 ≥ p2 ≥ ... ≥ pn ≥ 0).. The price when printing no less than si but less than si+1 pages is pi cents per page (for i=1..n-1). The price when printing no less than sn pages is pn cents per page. The third line containing m integers q1 .. qm (0 ≤ qi ≤ 109 ) are the queries.
Output
For each query qi, you should output the minimum amount of money (in cents) to pay if you want to print qi pages, one output in one line.
Sample Input
1 2 3 0 20 100 10 0 99 100
Sample Output
0 1000 1000
由于范围越大花费越小，所以用ps从后面刷一遍，找到每一位到下一个满足的最小花费 用二分查找找到满足的区间，判断在当前区间或补充到下一区间哪个花费更小即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 100010;
typedef long long ll;
ll s[maxn], p[maxn];
ll ps[maxn];
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
{
scanf("%lld%lld", s + i, p + i);
ps[i] = s[i] * p[i];
}
ll minv = ps[n];
for (int i = n - 1; i >= 1; i--)
{
minv = min(minv, ps[i + 1]);
ps[i] = minv;
}
for (int i = 0; i < m; i++)
{
ll ans = 0;
int k;
scanf("%d", &k);
int tmp = lower_bound(s + 1, s + n + 1, k) - s;
if (s[tmp] != k) tmp -= 1;
if (tmp == n) ans = k*p[tmp];
else
ans = min(k*p[tmp], ps[tmp]);
printf("%lld
", ans);
}
}
}