SDKD 2017 Spring Team Training B G题 -The Debut Album or URAL - 2018 (dp 三维)
2017-05-02 11:10
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题目描述
Pop-group “Pink elephant” entered on recording their debut album. In fact they have only two songs: “My love” and “I miss you”, but each of them has a large number of remixes.
The producer of the group said that the album should consist of n remixes. On second thoughts the musicians decided that the album will be of interest only if there are no more than a remixes on “My love” in
a row and no more than b remixes on “I miss you” in a row. Otherwise, there is a risk that even the most devoted fans won’t listen to the disk up to the end.
How many different variants to record the album of interest from n remixes exist? A variant is a sequence of integers 1 and 2, where ones denote remixes on “My love” and twos denote remixes on “I miss you”. Two variants
are considered different if for some i in one variant at i-th place stands one and in another variant at the same place stands two.
Input
The only line contains integers n, a, b (1 ≤ a, b ≤ 300; max( a, b) + 1 ≤ n ≤ 50 000).
Output
Output the number of different record variants modulo 10 9+7.
Example
题意: 连续个1 不超过a个 连续个2 不超过b个 问n个的话 有多少种
dp 推 状态方程 一直是弱项, 推了好久不出来, 用一个三维数组dp
思路:
dp[i][j][k] i代表构造第i个时 j代表是 1 还是2 k 是连续的长度
为了方便 用 i&1 操作 出现奇数或偶数 奇数1 偶数0 两种
4000
分别代表 1 和2
所以状态转移方程为
**************************************************************************************************
连续个1 不超过a 连续个2 不超过b 用0 代替1 用1 代替2
状态转移方程
i:1~n
j:1~a// 0时
dp[ i ][0][ j ] += dp[ i-1 ][0][ j-1 ] 当前与上一个都是0 则连续个数k的长度从j-1 变成j
dp[ i ][1][1] += dp[ i-1 ][0][ j ] 当前与上一个不一样, 连续个数k重置 为1
j:1~b// 1 时
dp[ i ][1][ j ] += dp[ i-1 ][1][ j-1 ] 同理
dp[ i ][0][1 ] += dp[ i-1 ][1][ j ]
**************************************************************************************************
数据量n: 5w a,b -300 三维的话 应该开 5w*2*300 - - 超内存了 ,直接爆了
所以这个地方又用到了滚动数组 滚啊滚就不会超了....
开个2*2*300的 就足够了
代码:
Pop-group “Pink elephant” entered on recording their debut album. In fact they have only two songs: “My love” and “I miss you”, but each of them has a large number of remixes.
The producer of the group said that the album should consist of n remixes. On second thoughts the musicians decided that the album will be of interest only if there are no more than a remixes on “My love” in
a row and no more than b remixes on “I miss you” in a row. Otherwise, there is a risk that even the most devoted fans won’t listen to the disk up to the end.
How many different variants to record the album of interest from n remixes exist? A variant is a sequence of integers 1 and 2, where ones denote remixes on “My love” and twos denote remixes on “I miss you”. Two variants
are considered different if for some i in one variant at i-th place stands one and in another variant at the same place stands two.
Input
The only line contains integers n, a, b (1 ≤ a, b ≤ 300; max( a, b) + 1 ≤ n ≤ 50 000).
Output
Output the number of different record variants modulo 10 9+7.
Example
input | output |
---|---|
3 2 1 | 4 |
Notes
In the example there are the following record variants: 112, 121, 211, 212.题意: 连续个1 不超过a个 连续个2 不超过b个 问n个的话 有多少种
dp 推 状态方程 一直是弱项, 推了好久不出来, 用一个三维数组dp
思路:
dp[i][j][k] i代表构造第i个时 j代表是 1 还是2 k 是连续的长度
为了方便 用 i&1 操作 出现奇数或偶数 奇数1 偶数0 两种
4000
分别代表 1 和2
所以状态转移方程为
**************************************************************************************************
连续个1 不超过a 连续个2 不超过b 用0 代替1 用1 代替2
状态转移方程
i:1~n
j:1~a// 0时
dp[ i ][0][ j ] += dp[ i-1 ][0][ j-1 ] 当前与上一个都是0 则连续个数k的长度从j-1 变成j
dp[ i ][1][1] += dp[ i-1 ][0][ j ] 当前与上一个不一样, 连续个数k重置 为1
j:1~b// 1 时
dp[ i ][1][ j ] += dp[ i-1 ][1][ j-1 ] 同理
dp[ i ][0][1 ] += dp[ i-1 ][1][ j ]
**************************************************************************************************
数据量n: 5w a,b -300 三维的话 应该开 5w*2*300 - - 超内存了 ,直接爆了
所以这个地方又用到了滚动数组 滚啊滚就不会超了....
开个2*2*300的 就足够了
代码:
#include <iostream> #include <cstring> #include <algorithm> #include <stdio.h> /* 连续个1 不超过a 连续个2 不超过b 用0 代替1 用1 代替2 状态转移方程 i:1~n j:1~a// 0时 dp[ i ][0][ j ] += dp[ i-1 ][0][ j-1 ] 当前与上一个都是0 则连续个数k的长度从j-1 变成j dp[ i ][1][1] += dp[ i-1 ][0][ j ] 当前与上一个不一样, 连续个数k重置 为1 j:1~b// 1 时 dp[ i ][1][ j ] += dp[ i-1 ][1][ j-1 ] 同理 dp[ i ][0][1 ] += dp[ i-1 ][1][ j ] */ const int mod=1e9+7; using namespace std; int dp[2][2][310]; int main() { int n,a,b; int i,j; while(cin>>n>>a>>b) { memset(dp,0,sizeof(dp)); dp[0][0][0]=dp[0][1][0]=1;//初始化 for(i=1;i<=n;i++) { int now=i&1;// 0/1 int pre=(now+1)&1; memset(dp[now],0,sizeof(dp[now])); for(j=1;j<=a;j++)//1 { dp[now][0][j]+=dp[pre][0][j-1];//当上一个现在一样的话 j为连续的 dp[now][0][1]%=mod; dp[now][1][1]+=dp[pre][0][j];//不一样是 后缀长度就变成1 重新开始 dp[now][1][1]%=mod; } for(j=1;j<=b;j++)//2 { dp[now][1][j]+=dp[pre][1][j-1]; dp[now][1][j]%=mod; dp[now][0][1]+=dp[pre][1][j]; dp[now][0][1]%=mod; } } int ans=0; for(i=1;i<=a;i++) { ans+=dp[n&1][0][i]; ans%=mod; } for(i=1;i<=b;i++) { ans+=dp[n&1][1][i]; ans%=mod; } printf("%d\n",ans); } return 0; }
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