題目鏈接：
HDU:http://acm.hdu.edu.cn/showproblem.php?pid=4791
ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5072
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
Source
2013 Asia Changsha Regional Contest
題意：
打印紙張，隨著張數的添加，每張的價格非遞增，給出 m 個詢問打印的張數，求出最小的花費。
PS:
保留打印a[i]份分別須要的錢，從后往前掃一遍，保證取得最優解。
然后再找到所打印的張數的區間，與沒有多打印，僅僅打印m張所需的花費比較！
HDU代碼例如以下：
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <iostream>
#define INF 1e18
using 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%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])//最后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;
}ZJU代碼例如以下：
#include <cstdio>
#include <algorithm>
#include <iostream>
#define INF 1e18
using namespace std;const int maxn = 100017;
typedef long 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%d",&n,&m);for(int i = 0; i < n; i++){scanf("%lld%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])//最后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;
}
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