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 i th 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?
Yuta has an array A of length n ，and the i th 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) .
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; }