hdu 5057 Argestes and Sequence（BestCoder Round #11)

Argestes and Sequence

                                                         Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                     Total Submission(s): 640    Accepted Submission(s): 163

Problem Description

Argestes has a lot of hobbies and likes solving query problems especially. One day Argestes came up with such a problem. You are given a sequence a consisting of N nonnegative integers, a[1],a[2],...,a[n].Then there are M operation on the sequence.An operation can be one of the following:
S X Y: you should set the value of a[x] to y(in other words perform an assignment a[x]=y).
Q L R D P: among [L, R], L and R are the index of the sequence, how many numbers that the Dth digit of the numbers is P.
Note: The 1st digit of a number is the least significant digit.

 

Input

In the first line there is an integer T , indicates the number of test cases.
For each case, the first line contains two numbers N and M.The second line contains N integers, separated by space: a[1],a[2],...,a[n]—initial value of array elements.
Each of the next M lines begins with a character type.
If type==S,there will be two integers more in the line: X,Y.
If type==Q,there will be four integers more in the line: L R D P.

[Technical Specification]
1<=T<= 50
1<=N, M<=100000
0<=a[i]<=$2^{31}$ - 1
1<=X<=N
0<=Y<=$2^{31}$ - 1
1<=L<=R<=N
1<=D<=10
0<=P<=9

 

Output

For each operation Q, output a line contains the answer.

 

Sample Input

15 710 11 12 13 14Q 1 5 2 1Q 1 5 1 0Q 1 5 1 1Q 1 5 3 0Q 1 5 3 1S 1 100Q 1 5 3 1

 

Sample Output

511501

 

以为是很简单的树状数组，开了个三维的，直接MLE了，可以离线处理，通过枚举每一位降低空间消耗。

代码：

//1296ms#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn=100000+1000;int a[11][maxn];int b[maxn];int c[maxn];int ans[maxn];int n,m;struct node{    int kind;    int l,r,d,p,pos,value;}op[maxn];int low(int k){    return k&(-k);}void update(int x,int k,int v){    while(k<maxn)    {        a[x][k]+=v;        k+=low(k);    }}int getsum(int x,int k){    int ans=0;    while(k>0)    {        ans+=a[x][k];        k-=low(k);    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)        {            scanf("%d",&b[i]);            c[i]=b[i];        }        char s[10];        for(int i=0;i<m;i++)        {            scanf("%s",s);            if(s[0]=='Q')            {                op[i].kind=0;                scanf("%d%d%d%d",&op[i].l,&op[i].r,&op[i].d,&op[i].p);            }            else            {                op[i].kind=1;                scanf("%d%d",&op[i].pos,&op[i].value);            }        }        int po=1;        for(int i=1;i<=10;i++)//按位枚举        {            memset(a,0,sizeof(a));            for(int j=1;j<=n;j++)            {                b[j]=c[j];//每次恢复为原来的值                int temp=(b[j]/po)%10;                update(temp,j,1);            }            for(int j=0;j<m;j++)            {                if(op[j].kind==1)//更改操作                {                    int temp=(b[op[j].pos]/po)%10;                    update(temp,op[j].pos,-1);                    temp=(op[j].value/po)%10;                    update(temp,op[j].pos,1);                    b[op[j].pos]=op[j].value;                }                else                {                    if(op[j].d==i)//对应位的查询                    ans[j]=getsum(op[j].p,op[j].r)-getsum(op[j].p,op[j].l-1);                    else                    ;                }            }            po=po*10;        }        for(int j=0;j<m;j++)        if(op[j].kind==0)        {            printf("%d\n",ans[j]);        }    }    return 0;}