Every month the stock manager
of The Royal Pearl prepares a list with the number of pearls needed in
each quality class. The pearls are bought on the local pearl market.
Each quality class has its own price per pearl, but for every complete
deal in a certain quality class one has to pay an extra amount of money
equal to ten pearls in that class. This is to prevent tourists from
buying just one pearl.
Also The Royal Pearl is suffering from the
slow-down of the global economy. Therefore the company needs to be more
efficient. The CFO (chief financial officer) has discovered that he can
sometimes save money by buying pearls in a higher quality class than is
actually needed. No customer will blame The Royal Pearl for putting
better pearls in the bracelets, as long as the prices remain the same.
For
example 5 pearls are needed in the 10 Euro category and 100 pearls are
needed in the 20 Euro category. That will normally cost: (5+10)*10 +
(100+10)*20 = 2350 Euro.
Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.
The
problem is that it requires a lot of computing work before the CFO
knows how many pearls can best be bought in a higher quality class. You
are asked to help The Royal Pearl with a computer program.
Given a
list with the number of pearls and the price per pearl in different
quality classes, give the lowest possible price needed to buy everything
on the list. Pearls can be bought in the requested, or in a higher
quality class, but not in a lower one.
first line of the input contains the number of test cases. Each test
case starts with a line containing the number of categories c (1 <= c
<= 100). Then, c lines follow, each with two numbers ai and pi. The
first of these numbers is the number of pearls ai needed in a class (1
<= ai <= 1000). The second number is the price per pearl pi in
that class (1 <= pi <= 1000). The qualities of the classes (and so
the prices) are given in ascending order. All numbers in the input are
integers.
2
100 1
100 2
3
1 10
1 11
100 12
思路：
解法一：（WA但数据测试正确）
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std; int T;
int number;
struct P{
__int64 v;
__int64 c;
}pearl[];
int exits[];
__int64 sum; int main()
{
scanf("%d",&T);
while(T--)
{
sum = ;
scanf("%d",&number);
for(int i = ;i <= number;i++){
scanf("%I64d%I64d",&pearl[i].c,&pearl[i].v);
exits[i] = ;
}
for(int i = ;i <= number;i++)
if(pearl[i-].c*pearl[i].v < (pearl[i-].c+)*pearl[i-].v){
pearl[i].c += pearl[i-].c;
exits[i-] = ;
}
for(int i = ;i <= number;i++)
if(exits[i]) sum += (pearl[i].c+)*pearl[i].v;
printf("%I64d\n",sum);
}
return ;
}
解法二（AC）：
#include <iostream>
#include <cstdio>
#include <cstring>
#define INF 0x7fffffff
using namespace std; int min(int a,int b)
{
return a<b?a:b;
} int main()
{
int T;
int sum[];
int val[];
int dp[];
int num[];
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
sum[] = ;
for(int i = ;i <= n;i++) {
scanf("%d%d",&num[i],&val[i]);
sum[i] = sum[i-]+num[i];
}
for(int i = ;i <= n;i++)
dp[i] = INF;
sum[] = ;
dp[] = ;
for(int i = ;i <= n;i++)
for(int j = ;j < i;j++)
dp[i] = min(dp[i],dp[j]+(sum[i]-sum[j]+)*val[i]);
printf("%d\n",dp[n]);
}
return ;
}