HDU 4791 & zoj 3726 Alice's Print Service (Math Table)

Question link:

HDU: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4791

Zju: http://acm.zju.edu.cn/onlinejudge/showProblem.do? Problemid = 5072

Problem descriptionalice 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. inputthe 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. outputfor each query Qi, you shocould output the minimum amount of money (in cents) to pay if you want to print Qi pages, one output in one line. sample Input

12 30 20 100 100 99 100

Sample output

010001000

Source2013 Asia Changsha Regional Contest

The HDU code is as follows:

# Include <cstdio> # include <cstdlib> # include <algorithm> # include <vector> # include <iostream> # define INF 1e18using namespace STD; const int maxn = 100017; typedef _ int64 ll; ll s [maxn], p [maxn], C [maxn]; int main () {int t; int n, m; ll tt; scanf ("% d", & T); While (t --) {scanf ("% d", & N, & M); For (INT I = 0; I <n; I ++) {scanf ("% i64d % i64d", & S [I], & P [I]);} ll Minn = inf; for (INT I = n-1; I> = 0; I --) {Minn = min (s [I] * P [I], Minn ); c [I] = Minn ;}for (INT I = 0; I <m; I ++) {scanf ("% i64d", & TT ); if (tt> = s [n-1]) // final printf ("% i64d \ n", TT * P [n-1]); else {int Pos = upper_bound (S, S + N, TT)-S; ll ans = TT * P [pos-1]; ans = min (ANS, C [POS]); printf ("% i64d \ n ", ans) ;}}return 0 ;}

The zju code is as follows:

# Include <cstdio> # include <algorithm> # include <iostream> # define INF 1e18using namespace STD; const int maxn = 100017; typedef long ll; ll s [maxn], P [maxn], C [maxn]; int main () {int t; int n, m; ll tt; scanf ("% d", & T ); while (t --) {scanf ("% d", & N, & M); For (INT I = 0; I <n; I ++) {scanf ("% LLD", & S [I], & P [I]);} ll Minn = inf; For (INT I = n-1; i> = 0; I --) {Minn = min (s [I] * P [I], Minn); C [I] = Minn ;} for (INT I = 0; I <m; I ++) {scanf ("% LLD", & TT); If (tt> = s [n-1]) // finally printf ("% LLD \ n", TT * P [n-1]); else {int Pos = upper_bound (S, S + N, TT)-S; ll ans = TT * P [pos-1]; ans = min (ANS, C [POS]); printf ("% LLD \ n", ANS );}}} return 0 ;}

HDU 4791 & zoj 3726 Alice's Print Service (Math Table)