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The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N‘s!
For example, given a set of coupons {1 2 4 -1}, and a set of product values {7 6 -2 -3} (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.
Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.
Input Specification:
Each input file contains one test case. For each case, the first line contains the number of coupons NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 105, and it is guaranteed that all the numbers will not exceed 230.
Output Specification:
For each test case, simply print in a line the maximum amount of money you can get back.
Sample Input:4 1 2 4 -1 4 7 6 -2 -3Sample Output:
43
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 using namespace std; 5 bool cmp(int a,int b){ 6 return a>b; 7 } 8 int main(){ 9 //freopen("D:\\INPUT.txt", "r", stdin); 10 int nc,np,i,j; 11 scanf("%d",&nc); 12 long long *ncp=new long long[nc+5]; 13 for(i=0;i<nc;i++){ 14 scanf("%lld",&ncp[i]); 15 } 16 sort(ncp,ncp+nc,cmp); 17 scanf("%d",&np); 18 long long *npp=new long long[np+5]; 19 for(i=0;i<np;i++){ 20 scanf("%lld",&npp[i]); 21 } 22 sort(npp,npp+np,cmp); 23 long long sum=0; 24 i=0,j=0; 25 while(i<nc&&j<np){ 26 long long mul=ncp[i]*npp[j]; 27 if(mul>=0){ 28 sum+=mul; 29 i++; 30 j++; 31 continue; 32 } 33 if(ncp[i]<0){ 34 break; 35 } 36 if(npp[j]<0){ 37 break; 38 } 39 } 40 //跳出循环的可能： 41 //i或j越界 42 //i指向一个负数 43 //j指向一个负数 44 int ii=nc-1,jj=np-1; 45 for(;ii>=i&&jj>=j&&ncp[ii]*npp[jj]>=0;ii--,jj--){ 46 sum+=ncp[ii]*npp[jj]; 47 } 48 printf("%lld\n",sum); 49 return 0; 50 }
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原文地址：http://www.cnblogs.com/Deribs4/p/4769120.html
