hdu5432Rikka with Array (数位dp+十进制转化为二进制)

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has an array A of
length n，and
the ith
element of A is
equal to the sum of all digits of i in
binary representation. For example,A[1]=1,A[3]=2,A[10]=2.

Now, Yuta wants to know the number of the pairs (i,j)(1≤i<j≤n) which
satisfy A[i]>A[j].

It is too difficult for Rikka. Can you help her?

 

Input

The first line contains a number T(T≤10)——The
number of the testcases.

For each testcase, the first line contains a number n(n≤10300).

 

Output

For each testcase, print a single number. The answer may be very large, so you only need to print the answer modulo 998244353.

 

Sample Input

1
10

 

Sample Output

7

When $n=10$, $A$ is equal to $1,1,2,1,2,2,3,1,2,2$.

So the answer is $7$.

题意：定义A[i]为i化为二进制后的1的个数，给你一个数n，你要在1~n之间找到两个数a,b，使得a<b且A[a]>A[b].问总共有多少这样的对数.
思路：这题和普通的数位dp不同，普通的数位dp是求单个数内的方案数，但是这道题是求对数，所以不能用之前的方法。这道题中我们要在记忆化搜索中同时枚举两个数，一个为大数，一个为小数，然后用dp[pos][cha][flag][same][lim]表示前pos位，大小两数化为二进制后1的个数差的绝对值为cha，flag表示是否小数化为二进制后的1的个数大于大数，same表示pos位前两个数是不是完全相同，lim表示大数是不是还有限制。据说题解只开了三维，感觉三维的条件太少了，我只会写五维的。。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 1005
#define MOD 998244353
char s[maxn];
int dp[maxn][maxn][2][2][2]; //dp[pos][cha][flag1][same][lim]表示前pos位，大小数1的个数差的绝对值是cha,小数中1的个数是不是大于大数中的个数
int wei[maxn],num[maxn];

int dfs(int pos,int cha,int flag,int same,int lim) //same表示大数和小数是不是前几位都一样
{
int i,j;
if(pos==0){
if(flag==1)return 1;
return 0;
}
if(dp[pos][cha][flag][same][lim]!=-1 ){
return dp[pos][cha][flag][same][lim];
}
int ans=0;
if(same==1){
if(lim==1){
if(wei[pos]==0){
ans=ans+dfs(pos-1,0,0,1,1); //都取0
if(ans>=MOD)ans-=MOD;
}
else if(wei[pos]==1){
ans=ans+dfs(pos-1,0,0,1,1);if(ans>=MOD)ans-=MOD; //都取1
ans=ans+dfs(pos-1,0,0,1,0);if(ans>=MOD)ans-=MOD; //都取0
ans=ans+dfs(pos-1,1,0,0,1);if(ans>=MOD)ans-=MOD; //大数取1，小数取0
}
}
else if(lim==0){
ans=ans+dfs(pos-1,0,0,1,0);if(ans>=MOD)ans-=MOD;
ans=ans+dfs(pos-1,0,0,1,0);if(ans>=MOD)ans-=MOD;
ans=ans+dfs(pos-1,1,0,0,0);if(ans>=MOD)ans-=MOD;
}
}
else if(same==0){
if(lim==1){
if(wei[pos]==1){
ans=ans+dfs(pos-1,cha,flag,same,0);if(ans>=MOD)ans-=MOD; //都加0
ans=ans+dfs(pos-1,cha,flag,same,1);if(ans>=MOD)ans-=MOD; //都加1
if(cha==0){
ans=ans+dfs(pos-1,1,0,same,1);if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,1,1,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
else if(cha>0){
if(flag==1){
int flag1;
if(cha==1)flag1=0;
else flag1=1;
ans=ans+dfs(pos-1,cha-1,flag1,same,1);if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,cha+1,1,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
else if(flag==0){
ans=ans+dfs(pos-1,cha+1,0,same,1 );if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,cha-1,0,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
}
}
else if(wei[pos]==0){
ans=ans+dfs(pos-1,cha,flag,same,1);if(ans>=MOD)ans-=MOD; //都加0
if(cha==0){
ans=ans+dfs(pos-1,cha+1,1,same,1);if(ans>=MOD)ans-=MOD; //大数加0，小数加1
}
else if(cha>0){
if(flag==1){
ans=ans+dfs(pos-1,cha+1,1,same,1);if(ans>=MOD)ans-=MOD;
}
else if(flag==0){
ans=ans+dfs(pos-1,cha-1,0,same,1);if(ans>=MOD)ans-=MOD;
}
}
}
}
else if(lim==0){
ans=ans+dfs(pos-1,cha,flag,same,0);if(ans>=MOD)ans-=MOD; //都加0
ans=ans+dfs(pos-1,cha,flag,same,0);if(ans>=MOD)ans-=MOD; //都加1
if(cha==0){
ans=ans+dfs(pos-1,1,0,same,0);if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,1,1,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
else if(cha>0){
if(flag==1){
int flag1;
if(cha==1)flag1=0;
else flag1=1;
ans=ans+dfs(pos-1,cha-1,flag1,same,0 );if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,cha+1,1,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
else if(flag==0){
ans=ans+dfs(pos-1,cha+1,0,same,0 );if(ans>=MOD)ans-=MOD; //大数加1
ans=ans+dfs(pos-1,cha-1,0,same,0);if(ans>=MOD)ans-=MOD; //小数加1
}
}
}
}
dp[pos][cha][flag][same][lim]=ans;
return ans;
}
void solve()
{
int i,j;
int len=strlen(s+1);
for(i=len;i>=1;i--){
num[i]=s[len+1-i]-'0';
}
int tot=0;
while(len){
for(i=len-1;i>=1;i--){
num[i]+=(num[i+1]&1)*10;
num[i+1]>>=1;
}
if(num[1]&1)wei[++tot]=1;
else wei[++tot]=0;
num[1]>>=1;
if(num[len]==0)len--;
}

memset(dp,-1,sizeof(dp)); //这里要注意，必须每一个样例都要初始化一遍，因为不同的数，dp[pos][cha][flag1][same][lim]的意义不同
printf("%d\n",dfs(tot,0,0,1,1)%MOD);
}

int main()
{
int n,m,i,j,T;
scanf("%d",&T);
while(T--)
{
scanf("%s",s+1);
solve();
}
return 0;
}