shu7-19【背包和母函数练习】
题目:http://acm.hdu.edu.cn/diy/contest_show.php?cid=20083
密码:shuacm
感觉他们学校的新生训练出的比较好。
今天很多题目都是强化了背包的转化。
关于背包转化成求最优解见分析:点击打开链接
贴个背包的模板:
//0-1背包, 代价为 cost, 获得的价值为 weight
// 每种物品最多只可以选一次
void ZeroOnePack(int cost, int weight)
{
for(int i = nValue; i >= cost; i--)
dp[i] = dp[i] + dp[i-cost]+weight;
} // 完全背包,代价为 cost, 获得的价值为 weight
// 每种物品可以选无限次, 或者在可选的有限次内能够装满背包
void CompletePack(int cost, int weight)
{
for(int i = cost; i <= nValue; i++)
dp[i] = dp[i] + dp[i-cost]+weight;
} //多重背包
//每种物品可以选有限次
void MultiplePack(int cost, int weight, int amount)
{
if(cost*amount >= nValue) CompletePack(cost, weight); else
{
int k = 1;
while(k < amount)
{
ZeroOnePack(k*cost, k*weight);
amount -= k;
k <<= 1;
}
ZeroOnePack(amount*cost, amount*weight);
}
}
关于转化后,除掉 weight,把 max 换成 sum 即可,具体分析见上面的博客
void ZeroOnePack(int cost)
{
for(int i = nValue; i >= cost; i--)
dp[i] = dp[i] + dp[i-cost];
} void CompletePack(int cost)
{
for(int i = cost; i <= nValue; i++)
dp[i] = dp[i] + dp[i-cost];
} void MultiplePack(int cost, int amount)
{
if(cost*amount >= nValue) CompletePack(cost); else
{
int k = 1;
while(k < amount)
{
ZeroOnePack(k*cost);
amount -= k;
k <<= 1;
}
ZeroOnePack(amount*cost);
}
}
A
和上面分析博客一样有木有
Problem A
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 40 Accepted Submission(s) : 32
Font: Times New Roman | Verdana | Georgia
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Problem Description
"The second problem is, given an positive integer N, we define an equation like this:
N=a[1]+a[2]+a[3]+...+a[m];
a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
4 = 4;
4 = 3 + 1;
4 = 2 + 2;
4 = 2 + 1 + 1;
4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"
Input
Output
Sample Input
4
10
20
Sample Output
5
42
627
#include<stdio.h>
#include<string.h> const int maxn = 150;
int dp[maxn]; int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
memset(dp,0,sizeof(dp));
dp[0] = 1;
for(int i = 1; i <= n; i++)
for(int j = i; j <= n; j++)
dp[j] += dp[j-i];
printf("%d\n", dp[n]); }
return 0;
}
B:母函数,待看中
为毛背包连样例 都 没有 出。。。感觉和 F 一样啊
Problem B
Time Limit : 1000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 23 Accepted Submission(s) : 18
Font: Times New Roman | Verdana | Georgia
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Problem Description
Input
每组数据的第一行是两个整数n(1 <= n <= 40),k(1 <= k <= 8)。
接着有k行,每行有两个整数a(1 <= a <= 8),b(1 <= b <= 10),表示学分为a的课有b门。
Output
Sample Input
2
2 2
1 2
2 1
40 8
1 1
2 2
3 2
4 2
5 8
6 9
7 6
8 8
Sample Output
2
445
直接套用母函数模板,背包还是没有想出来。。。没道理啊
#include<stdio.h>
#include<string.h> const int maxn = 1000000;
int c1[maxn];
int c2[maxn]; int a[10];
int b[10]; int main()
{
int T;
int n,k;
scanf("%d", &T);
while(T--)
{
memset(c1, 0, sizeof(c1));
memset(c2, 0, sizeof(c2));
c1[0] = 1;
scanf("%d%d", &n,&k);
for(int i = 1; i <= k; i++)
{
scanf("%d%d", &a[i],&b[i]);
} for(int i = 1; i <= k; i++) //有几种学分, 第几层括号
{
for(int j = 0; j <= n; j++) //在原有的基础上遍历
{
for(int t = 0; t+j <= n && t <= a[i]*b[i]; t += a[i]) //遍历第 i 层括号的每一项
{//每种学分选的次数有限制,所以同时 k 也不能超过当前总的
c2[t+j] += c1[j];
}
} for(int j = 0; j <= n; j++)
{
c1[j] = c2[j];
c2[j] = 0;
}
}
printf("%d\n", c1[n]);
}
return 0;
}
C :背包总容量改为总价值的一半,尽量装满背包。所求的最大体积则是第二个答案
第一个答案 = 总价值-第一个答案
注意:<= 0非法输入,跳出。而不是 n == -1 退出。
Problem C
Time Limit : 10000/5000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 39 Accepted Submission(s) : 11
Font: Times New Roman | Verdana | Georgia
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Problem Description
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is
N (0<N<1000) kinds of facilities (different value, different kinds).
Input
facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.
Output
Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1
Sample Output
20 10
40 40
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std; const int maxn = 50*50*100+10;
int dp[maxn];
int nValue; int v[60];
int m[60]; void ZeroOnePack(int cost)
{
for(int i = nValue; i >= cost; i--)
dp[i] = dp[i] + dp[i-cost];
} void CompletePack(int cost)
{
for(int i = cost; i <= nValue; i++)
dp[i] = dp[i] + dp[i-cost];
} void MultiplePack(int cost, int amount)
{
if(cost*amount >= nValue) CompletePack(cost); else
{
int k = 1;
while(k < amount)
{
ZeroOnePack(k*cost);
amount -= k;
k <<= 1;
}
ZeroOnePack(amount*cost);
}
} int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
if(n <= 0) break;
int sum = 0; for(int i = 0; i < n; i++)
{
scanf("%d%d", &v[i], &m[i]);
sum += v[i]*m[i];
}
nValue = sum/2; memset(dp,0,sizeof(dp));
dp[0] = 1; for(int i = 0; i < n; i++)
{
MultiplePack(v[i], m[i]);
} int ans1, ans2;
for(int i = nValue; i >= 0 ; i--)
{
if(dp[i] != 0)
{
ans2 = i;
break;
}
}
ans1 = sum-ans2;
printf("%d %d\n", ans1, ans2);
} return 0;
}
D:完全背包
Problem D
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 25 Accepted Submission(s) : 20
Font: Times New Roman | Verdana | Georgia
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Problem Description
available in Silverland.
There are four combinations of coins to pay ten credits:
ten 1-credit coins,
one 4-credit coin and six 1-credit coins,
two 4-credit coins and two 1-credit coins, and
one 9-credit coin and one 1-credit coin.
Your mission is to count the number of ways to pay a given amount using coins of Silverland.
Input
Output
Sample Input
2
10
30
0
Sample Output
1
4
27
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; int dp[310];
int a[20];
int nValue; void CompletePack(int cost)
{
for(int i = cost; i <= nValue; i++)
dp[i] = dp[i] + dp[i-cost];
} int main()
{
while(scanf("%d", &nValue) != EOF)
{
if(nValue == 0) break;
for(int i = 1; i <= 17; i++)
a[i] = i*i;
memset(dp,0,sizeof(dp));
dp[0] = 1; for(int i = 1; i <= 17; i++)
CompletePack(a[i]);
printf("%d\n", dp[nValue]);
}
}
E:不会
Problem E
Time Limit : 30000/15000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 4 Accepted Submission(s) : 1
Font: Times New Roman | Verdana | Georgia
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Problem Description
select some stones and arrange them in line to form a beautiful pattern. After several arrangements he finds it very hard for him to enumerate all the patterns. So he asks you to write a program to count the number of different possible patterns. Two patterns
are considered different, if and only if they have different number of stones or have different colors on at least one position.
Input
available stones of each color respectively. All the input numbers will be nonnegative and no more than 100.
Output
which is a prime number.
Sample Input
3
1 1 1
2
1 2
Sample Output
Case 1: 15
Case 2: 8
Hint
BGM; BMG; GBM; GMB; MBG; MGB.
F:多重背包求解
Problem F
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 79 Accepted Submission(s) : 23
Font: Times New Roman | Verdana | Georgia
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Problem Description
“Oh, God! How terrible! ”

Don’t be so afraid, guys. Although he hides in a cave of Hang Zhou, he dares not to go out. Laden is so bored recent years that he fling himself into some math problems, and he said that if anyone can solve his problem, he will give himself up!
Ha-ha! Obviously, Laden is too proud of his intelligence! But, what is his problem?
“Given some Chinese Coins (硬币) (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value that you cannot pay with given coins.”
You, super ACMer, should solve the problem easily, and don’t forget to take $25000000 from Bush!
Input
Output
Sample Input
1 1 3
0 0 0
Sample Output
4
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std; const int maxn = 8000+10;
int dp[maxn];
int nValue; void ZeroOnePack(int cost)
{
for(int i = nValue; i >= cost; i--)
dp[i] = dp[i] + dp[i-cost];
} void CompletePack(int cost)
{
for(int i = cost; i <= nValue; i++)
dp[i] = dp[i] + dp[i-cost];
} void MultiplePack(int cost, int amount)
{
if(cost*amount >= nValue) CompletePack(cost); else
{
int k = 1;
while(k < amount)
{
ZeroOnePack(k*cost);
amount -= k;
k <<= 1;
}
ZeroOnePack(amount*cost);
}
} int main()
{
int a,b,c;
while(scanf("%d%d%d", &a,&b,&c) != EOF)
{
if(a == 0 && b == 0 && c == 0) break; memset(dp,0,sizeof(dp));
dp[0] = 1;
nValue = 1*a+2*b+5*c; MultiplePack(1, a);
MultiplePack(2, b);
MultiplePack(5, c); for(int i = 0; i < maxn; i++)
{
if(dp[i] == 0)
{
printf("%d\n", i);
break;
}
}
}
return 0;
}
G:签到。。。
Problem G
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 45 Accepted Submission(s) : 40
Font: Times New Roman | Verdana | Georgia
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Problem Description
什么问题?他研究的问题是蟠桃一共有多少个!
不过,到最后,他还是没能解决这个难题,呵呵^-^
当时的情况是这样的:
第一天悟空吃掉桃子总数一半多一个,第二天又将剩下的桃子吃掉一半多一个,以后每天吃掉前一天剩下的一半多一个,到第n天准备吃的时候只剩下一个桃子。聪明的你,请帮悟空算一下,他第一天开始吃的时候桃子一共有多少个呢?
Input
Output
Sample Input
2
4
Sample Output
4
22
#include<stdio.h> int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
int ans = 1;
while(--n)
{
ans = (ans+1)*2;
}
printf("%d\n", ans);
}
return 0;
}
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