正文
SDUT 2408:Pick apples
小程序:扫一扫查出行
【扫一扫了解最新限行尾号】
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【扫一扫了解最新限行尾号】
复制小程序
Pick applesTime Limit: 1000MS Memory limit: 165536K
题目描述Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she has never seen so many apples before. Each kind
of apple has a size and a price to be sold. Now the little girl wants to gain more profits, but she does not know how. So she asks you for help, and tell she the most profits she can gain.
输入In the first line there is an integer T (T <= 50), indicates the number of test cases.
In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S<= 100) and the price (1 <= P <= 10000) of this kind of apple.
In the fourth line there is an integer V,(1 <= V <= 100,000,000)indicates the volume of the girl's bag.
输出For each case, first output the case number then follow the most profits she can gain.
示例输入1
1 1
2 1
3 1
6
示例输出Case 1: 6
可以考虑贪心+背包的思想,贪心优先!以防止超时。
#include<stdio.h>
#define max(a,b) (a>b?a:b)
int s[4],p[4],kk=1;
long long dp[4][1000005];
long long dpp(int v)
{
for(int i=1; i<=3; ++i)
for(int j=0; j<=v; ++j)
{
if(j>=s[i])dp[i][j]=max(dp[i-1][j],dp[i][j-s[i]]+p[i]);
else dp[i][j]=dp[i-1][j];
}
return dp[3][v];
}
int main()
{
int t;
scanf( "%d",&t );
while( t-- )
{
int maxx=0,v,g=1;
for(int i=1; i<=3; i++)
{
scanf("%d%d",&s[i],&p[i]);
g*=s[i];
}
scanf("%d",&v);
for(int i=1; i<=3; ++i)
maxx=maxx>g/s[i]*p[i]?maxx:g/s[i]*p[i];
long long ans=(long long)(v/g)*maxx;
v%=g;
ans+=dpp(v);
printf("Case %d: %lld\n",kk++,ans);
}
return 0;
}
Once ago, there is a mystery yard which only produces three kinds of apples. The number of each kind is infinite. A girl carrying a big bag comes into the yard. She is so surprised because she has never seen so many apples before. Each kind
of apple has a size and a price to be sold. Now the little girl wants to gain more profits, but she does not know how. So she asks you for help, and tell she the most profits she can gain.
输入In the first line there is an integer T (T <= 50), indicates the number of test cases.
In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S<= 100) and the price (1 <= P <= 10000) of this kind of apple.
In the fourth line there is an integer V,(1 <= V <= 100,000,000)indicates the volume of the girl's bag.
输出For each case, first output the case number then follow the most profits she can gain.
示例输入1
1 1
2 1
3 1
6
示例输出Case 1: 6
可以考虑贪心+背包的思想,贪心优先!以防止超时。
#include<stdio.h>
#define max(a,b) (a>b?a:b)
int s[4],p[4],kk=1;
long long dp[4][1000005];
long long dpp(int v)
{
for(int i=1; i<=3; ++i)
for(int j=0; j<=v; ++j)
{
if(j>=s[i])dp[i][j]=max(dp[i-1][j],dp[i][j-s[i]]+p[i]);
else dp[i][j]=dp[i-1][j];
}
return dp[3][v];
}
int main()
{
int t;
scanf( "%d",&t );
while( t-- )
{
int maxx=0,v,g=1;
for(int i=1; i<=3; i++)
{
scanf("%d%d",&s[i],&p[i]);
g*=s[i];
}
scanf("%d",&v);
for(int i=1; i<=3; ++i)
maxx=maxx>g/s[i]*p[i]?maxx:g/s[i]*p[i];
long long ans=(long long)(v/g)*maxx;
v%=g;
ans+=dpp(v);
printf("Case %d: %lld\n",kk++,ans);
}
return 0;
}
In the first line there is an integer T (T <= 50), indicates the number of test cases.
In each case, there are four lines. In the first three lines, there are two integers S and P in each line, which indicates the size (1 <= S<= 100) and the price (1 <= P <= 10000) of this kind of apple.
In the fourth line there is an integer V,(1 <= V <= 100,000,000)indicates the volume of the girl's bag.
For each case, first output the case number then follow the most profits she can gain.
示例输入1
1 1
2 1
3 1
6
示例输出Case 1: 6
可以考虑贪心+背包的思想,贪心优先!以防止超时。
#include<stdio.h>
#define max(a,b) (a>b?a:b)
int s[4],p[4],kk=1;
long long dp[4][1000005];
long long dpp(int v)
{
for(int i=1; i<=3; ++i)
for(int j=0; j<=v; ++j)
{
if(j>=s[i])dp[i][j]=max(dp[i-1][j],dp[i][j-s[i]]+p[i]);
else dp[i][j]=dp[i-1][j];
}
return dp[3][v];
}
int main()
{
int t;
scanf( "%d",&t );
while( t-- )
{
int maxx=0,v,g=1;
for(int i=1; i<=3; i++)
{
scanf("%d%d",&s[i],&p[i]);
g*=s[i];
}
scanf("%d",&v);
for(int i=1; i<=3; ++i)
maxx=maxx>g/s[i]*p[i]?maxx:g/s[i]*p[i];
long long ans=(long long)(v/g)*maxx;
v%=g;
ans+=dpp(v);
printf("Case %d: %lld\n",kk++,ans);
}
return 0;
}
1
1 1
2 1
3 1
6
Case 1: 6
可以考虑贪心+背包的思想,贪心优先!以防止超时。
#include<stdio.h>
#define max(a,b) (a>b?a:b)
int s[4],p[4],kk=1;
long long dp[4][1000005];
long long dpp(int v)
{
for(int i=1; i<=3; ++i)
for(int j=0; j<=v; ++j)
{
if(j>=s[i])dp[i][j]=max(dp[i-1][j],dp[i][j-s[i]]+p[i]);
else dp[i][j]=dp[i-1][j];
}
return dp[3][v];
}
int main()
{
int t;
scanf( "%d",&t );
while( t-- )
{
int maxx=0,v,g=1;
for(int i=1; i<=3; i++)
{
scanf("%d%d",&s[i],&p[i]);
g*=s[i];
}
scanf("%d",&v);
for(int i=1; i<=3; ++i)
maxx=maxx>g/s[i]*p[i]?maxx:g/s[i]*p[i];
long long ans=(long long)(v/g)*maxx;
v%=g;
ans+=dpp(v);
printf("Case %d: %lld\n",kk++,ans);
}
return 0;
}