数位dp HDU 3709 Balanced Number

Balanced Number

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1591    Accepted Submission(s): 696

Problem Description

 

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

 

 

Input

 

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10¹⁸).

 

 

Output

 

For each case, print the number of balanced numbers in the range [x, y] in a line.

 

 

Sample Input

 

20 97604 24324

 

 

Sample Output

 

10897

 

 

Author

 

GAO, Yuan

 

 

Source

 

2010 Asia Chengdu Regional Contest

 

 

 

题意：要你找出[x,y]的平衡数的数目。

 

思路：我们用 dp[i][j][k]表示i位数，对称中心为j，平衡情况为k的数有多少个，那么我们能得出递推式dp[i+1][j][k+(j-(i+1)*val]+=dp[i][j][k] 这个我用bfs来推的。因为k并不一定不是每一个都有，所以用for循环可能会有很多无用的状态。然后输出的话，套个dfs的模板就行了，最后要注意的是0被我们算了多次，要减去重复的。

 

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string.h>#include<queue>#include<algorithm>#include<cmath>using namespace std;const int maxn=2*2000+5;const int zero=2000;#define LL long longLL dp[20][20][maxn];char digit[20];int n;struct State{    int ps,pv,bal;    State(int ps,int pv,int bal)    :ps(ps),pv(pv),bal(bal) {}};bool vis[20][20][maxn];void pre_init(){    memset(dp,0,sizeof(dp));    queue<State> q;    for(int p=0;p<=19;p++) {        dp[0][p][zero]=1;        q.push(State(0,p,zero));    }    memset(vis,0,sizeof(vis));    while(q.size()) {        State tmp=q.front(); q.pop();        int i=tmp.ps,j=tmp.pv,k=tmp.bal;     //   printf("%d %d %d: %lld\n",i,j,k,dp[i][j][k]);        if(i==19) continue;        for(int val=0;val<=9;++val) {            int kk=k+(j-(i+1))*val;            dp[i+1][j][kk]+=dp[i][j][k];            if(!vis[i+1][j][kk]) q.push(State(i+1,j,kk));            vis[i+1][j][kk]=true;        }    }}LL dfs(int cur,int pivot,int bal,bool limit){    if(!limit || cur==0) return dp[cur][pivot][2*zero-bal];    int tail=limit?digit[cur]-'0':9;    LL ret=0;    for(int val=0;val<=tail;++val)        ret+=dfs(cur-1,pivot,bal+(pivot-cur)*val,limit&&val==tail);    return ret;}LL Cal_Num(){    LL ret=0;    for(int i=1;i<=n;++i)        ret+=dfs(n,i,zero,true);    return ret-(n-1);}void Test(){    int i,j,k;    while(cin>>i>>j>>k) {        k+=zero;        cout<<dp[i][j][k]<<endl;    }}int bal_cnt[100000];bool balance(int p,char * s,int n){    int val=0;    for(int i=1;i<=n;++i)        val+=(p-i)*(s[i]-'0');    return val==0;}bool is_bal(int x){    char s[1000]; sprintf(s+1,"%d",x);    n=strlen(s+1);    for(int pivot=1;pivot<=n;++pivot)        if(balance(pivot,s,n)) return true;    return false;}void Brute(){    bal_cnt[0]=1;    for(int i=1;i<100000;++i)    {        bal_cnt[i]=bal_cnt[i-1]+(is_bal(i));       // printf("%d: %d\n",i,bal_cnt[i]);    }    int x;    freopen("out.txt","w",stdout);    for(int i=0;i<100000;++i) printf("%d: %d\n",i,bal_cnt[i]);}void GetData(){    freopen("in.txt","w",stdout);    printf("1000000\n");    for(int i=0;i<1000000;++i) printf("0 %d\n",i);}int main(){   // GetData(); return 0;//   freopen("in.txt","r",stdin);//   freopen("out.txt","w",stdout);  //  Brute(); return 0;    pre_init();   // Test();    int T;cin>>T;    int cnt=0;    while(T--)    {        LL x; scanf("%I64d",&x);        LL ans=0;        if(x>0) {            --x;            sprintf(digit+1,"%I64d",x);            n=strlen(digit+1);            reverse(digit+1,digit+1+n);            ans=Cal_Num();        }        scanf("%s",digit+1);        n=strlen(digit+1);        reverse(digit+1,digit+1+n);        ans = Cal_Num()-ans;        printf("%I64d\n",ans);    }}