hdu5632 Rikka with Array 數位dp

Rikka with Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 301    Accepted Submission(s): 119

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 $7$.

 

Source

BestCoder Round #73 (div.2)

 

Recommend

hujie



題目描述：給定一個數n(0 < n < 10^300)，問有多少個數對(i , j)，滿足1<=i < j <= n且A[i] > A[j]，其中A[x]是x化成二進制之後中1的個數

思路：定義dp[len][sum][limit]表示當前枚舉到第len位，已經枚舉出的兩個數的數位中1的個數差爲i - j + 1000（加1000是因爲差可能爲負），枚舉到第len位時i和j的狀態爲limit時合法的數對，其中
               limit == 0      表示i<j < n
               limit == 1      表示i<j = n
               limit == 2      表示i= j < n
               limit == 3      表示i =j = n
之所以要定義這四種狀態，是因爲這四種狀態下在填i和j的第len-1位的時候的限制不同，轉移方程見代碼

收穫：1、對於兩個數的數位dp，其實道理和一個數時是一樣的，每次枚舉兩個數的這一位要放什麼數，只是貼上界更麻煩
            2、大整數轉二進制並不需要高精度，從別人代碼裏學來了一個辦法替掉了自己原本代碼裏的高精度

#pragma warning(disable:4786)
#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cmath>
#include<string>
#include<sstream>
#include<bitset>
#define LL long long
#define FOR(i,f_start,f_end) for(int i=f_start;i<=f_end;++i)
#define mem(a,x) memset(a,x,sizeof(a))
#define lson l,m,x<<1
#define rson m+1,r,x<<1|1
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const double PI = acos(-1.0);
const double eps=1e-6;
const int maxn = 1000;
const int base = 1000;
LL dp[4][maxn][maxn * 2] ;
char s[maxn];
int num[maxn] , bits[maxn];
LL dfs(int len , int sum , int limit , int zero)
{
    if(sum > base && sum - base > len - 1)      return 0;
    if(len == 1)        return (sum < base) && (!zero);
    if(dp[limit][len][sum] != -1)       return dp[limit][len][sum];
    LL res = 0;
    for(int i = 0 ; i < 2 ; i++){
        for(int j = 0 ; j < 2 ; j++){
            if(limit == 0){
                res += dfs(len - 1 , sum + i - j , 0 , zero && j == 0);
                res %= mod;
            }
            else if(limit == 1){
                if(i > bits[len - 1])     continue;
                res += dfs(len - 1 , sum + i - j , i == bits[len - 1] ? 1 : 0, zero && j == 0);
                res %= mod;
            }
            else if(limit == 2){
                if(i < j)       continue;
                res += dfs(len - 1 , sum + i - j , i == j ? 2 : 0, zero && j == 0);
                res %= mod;
            }
            else{
                if(i == bits[len - 1]){
                    if(j < i)       res += dfs(len - 1 , sum + i - j , 1, zero && j == 0);
                    if(j == i)     res += dfs(len - 1 , sum + i - j , 3, zero && j == 0);
                }
                else if(i < bits[len - 1]){
                    if(j < i)       res += dfs(len - 1 , sum + i - j , 0, zero && j == 0);
                    if(j == i)     res += dfs(len - 1 , sum + i - j , 2, zero && j == 0);
                }
                res %= mod;
            }
        }
    }
    dp[limit][len][sum] = res;
    return res;
}
//convert函數用於大十進制數轉二進制數，其中len是大整數的位數，大整數已經存在num數組裏了
//如大整數爲14236，則len = 5 , num[1] = 6 , num[2] = 3 , num[3] = 2 , num[4] = 4,num[5] = 1,num[0]只用於判斷
//設大整數爲n，大整數位數爲len，時間複雜度是（logn * len）
int convert(int len)
{
    int m = 1;
    while(len){
        for(int i = len ; i ; i--){
            num[i - 1] += (num[i] & 1) * 10;
            num[i] >>= 1;
        }   //這個操作一結束num中存的大整數就變成了原來的一半
        bits[m++] = (num[0] != 0);
        num[0] = 0;
        if(!num[len])
            --len;
    }
    return m;
}
LL solve(int len)
{
    int cnt = convert(len);
    LL ret = dfs(cnt , base , 3  , 1);
    return ret;
}
int main()
{
    int T;
    scanf("%d" , &T);
    mem(dp , -1);
    while(T--){
        mem(dp[1] , -1);
        mem(dp[3] , -1);
        scanf("%s" , s + 1);
        int len = strlen(s + 1);
        for(int i = 1 ; i<= len ; i++){
            num[i] = s[len - i + 1] - '0';
        }
        LL ans = solve(len);
        printf("%lld\n",ans);
    }
    return 0;
}