UVAlive 6611 Alice's Print Service 二分
2015-10-15 20:41
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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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
1
2 3
0 20 100 10
0 99 100
[align=left]Sample Output[/align]
0
1000
1000
[align=left]Source[/align]
2013 Asia Changsha Regional Contest
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.
[align=left]Input[/align]
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.
[align=left]Output[/align]
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.
[align=left]Sample Input[/align]
1
2 3
0 20 100 10
0 99 100
[align=left]Sample Output[/align]
0
1000
1000
[align=left]Source[/align]
2013 Asia Changsha Regional Contest
/* *********************************************** Author :pk28 Created Time :2015/10/15 20:05:30 File Name :changchun1.cpp ************************************************ */ #include <iostream> #include <cstring> #include <cstdlib> #include <stdio.h> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <iomanip> #include <list> #include <deque> #include <stack> #define ull unsigned long long #define ll long long #define mod 90001 #define INF 1e13 #define maxn 100070 #define cle(a) memset(a,0,sizeof(a)) const ull inf = 1LL << 61; const double eps=1e-5; using namespace std; bool cmp(int a,int b){ return a>b; } int n,m; ll x[maxn],p[maxn]; ll minn[maxn]; int main() { #ifndef ONLINE_JUDGE freopen("in.txt","r",stdin); #endif //freopen("out.txt","w",stdout); int t; cin>>t; while(t--){ cin>>n>>m; for(int i=1;i<=n;i++){ scanf("%lld %lld",&x[i],&p[i]); } minn[n+1]=INF; for(int i=n;i>=1;i--){ minn[i]=min(minn[i+1],p[i]*x[i]); } ll a,ans; for(int i=1;i<=m;i++){ scanf("%lld",&a); int pos=lower_bound(x+1,x+1+n,a)-x; ans=min(a*p[pos-1],minn[pos]); printf("%lld\n",ans); } } return 0; }
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