hdu5632 Rikka with Array 数位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

110

 

Sample Output

7When $n=10$, $A$ 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



题目描述：给定一个数n(0 < n < 10^300)，问有多少个数对(i , j)，满足1<=i < j <= n且A[i] > A[j]，其中A[x]是x化成二进制之后中1的个数

思路：定义dp[len][sum][limit]表示当前枚举到第len位，已经枚举出的两个数的数位中1的个数差为i - j + 1000（加1000是因为差可能为负），枚举到第len位时i和j的状态为limit时合法的数对，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
之所以要定义这四种状态，是因为这四种状态下在填i和j的第len-1位的时候的限制不同，转移方程见代码

收获：1、对于两个数的数位dp，其实道理和一个数时是一样的，每次枚举两个数的这一位要放什么数，只是贴上界更麻烦
            2、大整数转二进制并不需要高精度，从别人代码里学来了一个办法替掉了自己原本代码里的高精度

#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|1using 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函数用于大十进制数转二进制数，其中len是大整数的位数，大整数已经存在num数组里了//如大整数为14236，则len = 5 , num[1] = 6 , num[2] = 3 , num[3] = 2 , num[4] = 4,num[5] = 1,num[0]只用于判断//设大整数为n，大整数位数为len，时间复杂度是（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;        }   //这个操作一结束num中存的大整数就变成了原来的一半        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;}