hdu 4906 Our happy ending 状压dp
2017-08-06 15:25
375 查看
http://acm.hdu.edu.cn/showproblem.php?pid=4906
题意:一个数量为n的序列,每个数0<=a[i]<=l,要求序列中的某些数之和等于k。求有多少种赋值方案。
题解:样例:可以赋值0 2,1 2,2 2,1 1,2 1,2 0。dp[i]:i转化为二进制,第一位表示和为1,第二位表示和为2……,二进制1表示有,0表示无。如i=3,二进制是11,表示有和为1,2的存在。下一个状态:next=(1<<(p-1))|i|(i<<k)&size。1、表示和为p的存在,2、i本身,3、有和为(i+p)存在。
代码:
#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
const int N = 100000 + 5;
const int mod = 1000000000 + 7;
int t,n,k,l,minn,v;
LL dp[1<<21],ans;
int main(){
scanf("%d",&t);
while(t--){
int size,next;
scanf("%d%d%d",&n,&k,&l);
size=(1<<k)-1,minn=min(k,l),mem(dp,0),dp[0]=1,ans=0;
for(int i=1;i<=n;i++){
for(int j=size;j>=0;j--){
if(dp[j]){
v=dp[j];
for(int p=1;p<=minn;p++){
next=(1<<(p-1))|j|((j<<p)&size);
dp[next]=(dp[next]+v)%mod;
}
if(k<l)dp[j]=(dp[j]+(LL)v*(l-k))%mod;
}
}
}
for(int i=0;i<=size;i++){
if(i&(1<<(k-1))){
ans=(ans+dp[i])%mod;
}
}
printf("%lld\n",ans);
}
return 0;
}
Our happy endingTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1275 Accepted Submission(s): 447 Problem Description There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli. Y*wan still remember the day he first meets the devil. Now everything is done and the devil is gone. Y*wan feel very sad and suicide. You feel guilty after killing so many loli, so you suicide too. Nobody survive in this silly story, but there is still some hope, because this is just a silly background story during one programming contest! And the last problem is: Given a sequence a_1,a_2,...,a_n, if we can take some of them(each a_i can only be used once), and they sum to k, then we say this sequence is a good sequence. How many good sequence are there? Given that each a_i is an integer and 0<= a_i <= L. You should output the result modulo 10^9+7. Input The first line contains an integer T, denoting the number of the test cases. For each test case, the first line contains 3 integers n, k, L. T<=20, n,k<=20 , 0<=L<=10^9. Output For each cases, output the answer in a single line. Sample Input 1 2 2 2 Sample Output 6 Author WJMZBMR Source 2014 Multi-University Training Contest 4 |
题解:样例:可以赋值0 2,1 2,2 2,1 1,2 1,2 0。dp[i]:i转化为二进制,第一位表示和为1,第二位表示和为2……,二进制1表示有,0表示无。如i=3,二进制是11,表示有和为1,2的存在。下一个状态:next=(1<<(p-1))|i|(i<<k)&size。1、表示和为p的存在,2、i本身,3、有和为(i+p)存在。
代码:
#include<bits/stdc++.h>
#define debug cout<<"aaa"<<endl
#define mem(a,b) memset(a,b,sizeof(a))
#define LL long long
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define MIN_INT (-2147483647-1)
#define MAX_INT 2147483647
#define MAX_LL 9223372036854775807i64
#define MIN_LL (-9223372036854775807i64-1)
using namespace std;
const int N = 100000 + 5;
const int mod = 1000000000 + 7;
int t,n,k,l,minn,v;
LL dp[1<<21],ans;
int main(){
scanf("%d",&t);
while(t--){
int size,next;
scanf("%d%d%d",&n,&k,&l);
size=(1<<k)-1,minn=min(k,l),mem(dp,0),dp[0]=1,ans=0;
for(int i=1;i<=n;i++){
for(int j=size;j>=0;j--){
if(dp[j]){
v=dp[j];
for(int p=1;p<=minn;p++){
next=(1<<(p-1))|j|((j<<p)&size);
dp[next]=(dp[next]+v)%mod;
}
if(k<l)dp[j]=(dp[j]+(LL)v*(l-k))%mod;
}
}
}
for(int i=0;i<=size;i++){
if(i&(1<<(k-1))){
ans=(ans+dp[i])%mod;
}
}
printf("%lld\n",ans);
}
return 0;
}
Our happy endingTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1275 Accepted Submission(s): 447 Problem Description There is an old country and the king fell in love with a devil. The devil always asks the king to do some crazy things. Although the king used to be wise and beloved by his people. Now he is just like a boy in love and can’t refuse any request from the devil. Also, this devil is looking like a very cute Loli. Y*wan still remember the day he first meets the devil. Now everything is done and the devil is gone. Y*wan feel very sad and suicide. You feel guilty after killing so many loli, so you suicide too. Nobody survive in this silly story, but there is still some hope, because this is just a silly background story during one programming contest! And the last problem is: Given a sequence a_1,a_2,...,a_n, if we can take some of them(each a_i can only be used once), and they sum to k, then we say this sequence is a good sequence. How many good sequence are there? Given that each a_i is an integer and 0<= a_i <= L. You should output the result modulo 10^9+7. Input The first line contains an integer T, denoting the number of the test cases. For each test case, the first line contains 3 integers n, k, L. T<=20, n,k<=20 , 0<=L<=10^9. Output For each cases, output the answer in a single line. Sample Input 1 2 2 2 Sample Output 6 Author WJMZBMR Source 2014 Multi-University Training Contest 4 |
相关文章推荐
- hdu 4906 Our happy ending (多校第4场 )状压DP
- HDU 4906 Our happy ending (状压DP)
- hdu 4906 Our happy ending(状压dp)
- HDU 4906 Our happy ending 状压DP
- HDU 4906 Our happy ending (状压DP)
- HDU 4906 Our happy ending 状压DP
- hdu 4906 Our happy ending 状态压缩dp
- hdu 4906 Our happy ending 状态压缩dp 较难
- hdu 4906 Our happy ending (状态压缩dp)
- hdu 4906 Our happy ending 。神奇的状态转移方程,记录下
- hdu 4906 状压dp
- hdu 4906——Our happy ending
- HDU 4906 Our happy ending(2014 Multi-University Training Contest 4)
- HDU 4906 Our happy ending 解题报告(递推)
- HDU 3920 Clear All of Them I 状压DP
- 方格取数(1) HDU - 1565(状压dp)
- 【hdu】4352 XHXJ's LIS【状压+数位dp】
- HDU 3182 Hamburger Magi(状压dp)
- HDU 5418 Victor and World (状压DP入门)
- HDU 5163 状压DP