hdu 3049 Data Processing

传送门：这里。

Chinachen is a football fanatic, and his favorite football club is Juventus fc. In order to buy a ticket of Juv, he finds a part-time job in Professor Qu’s lab.And now, Chinachen have received an arduous task——Data Processing.The data was made up with N positive integer (n1, n2, n3, … ), he may calculate the number , you can assume mod N =0. Because the number is too big to count, so P mod 1000003 is instead.Chinachen is puzzled about it, and can’t find a good method to finish the mission, so he asked you to help him.

Input

The first line of input is a T, indicating the test cases number.There are two lines in each case. The first line of the case is an integer N, and N<=40000. The next line include N integer numbers n1,n2,n3… (ni<=N).

Output

For each test case, print a line containing the test case number ( beginning with 1) followed by the P mod 1000003.

Sample Input

231 1 341 2 1 4

Sample Output

Case 1:4Case 2:6

【题意】

给你长度为len的一个数组[a1,,,an]，要你求((2^a1+2^a2+…+2^an)/len)%mod。其中题目保证了分子是分母的整数倍。

【分析】

显然是求逆元的过程。以前只是会用板子，今天终于懂了逆元的作用。分析如下：

设x是len关于mod的逆元，那么显然有len*x==1.所以上式可以化简为：
((2^a1+2^a2+…+2^an)/len)%mod = ((2^a1+2^a2+…+2^an)/len*(len*x))%mod = ((2^a1+2^a2+…+2^an)*x)%mod

这样就把一个除法问题转化成了乘法问题，这就是逆元的巧妙之处。然后就是扩展欧几里得的事了，没啥好说的。

【代码】

#include<iostream>#include<cstdio>#include<cstring>#include<string.h>#include<algorithm>#include<vector>#include<cmath>#include<stdlib.h>#include<time.h>#include<stack>#include<set>#include<map>#include<queue>#include<sstream>using namespace std;#define rep0(i,l,r) for(int i = (l);i < (r);i++)#define rep1(i,l,r) for(int i = (l);i <= (r);i++)#define rep_0(i,r,l) for(int i = (r);i > (l);i--)#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)#define MS0(a) memset(a,0,sizeof(a))#define MS_1(a) memset(a,-1,sizeof(a))#define MSinf(a) memset(a,0x3f,sizeof(a))#define MSfalse(a) memset(a,false,sizeof(a))#define inf 0x3f3f3f3f#define lson i<<1,l,mid#define rson ((i<<1)|1),mid+1,r#define uint unsigned inttypedef pair<int,int> PII;#define A first#define B second#define pb push_back#define mp make_pair#define ll long long#define eps 1e-8inline void read1(int &num) {    char in;    bool IsN=false;    in=getchar();    while(in!='-'&&(in<'0'||in>'9')) in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else num=in-'0';    while(in=getchar(),in>='0'&&in<='9') {        num*=10,num+=in-'0';    }    if(IsN) num=-num;}inline void read2(int &a,int &b) {    read1(a);    read1(b);}inline void read3(int &a,int &b,int &c){    read1(a);    read1(b);    read1(c);}inline void read1(ll &num) {    char in;    bool IsN=false;    in=getchar();    while(in!='-'&&(in<'0'||in>'9')) in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else num=in-'0';    while(in=getchar(),in>='0'&&in<='9') {        num*=10,num+=in-'0';    }    if(IsN) num=-num;}inline void read2(ll &a,ll &b) {    read1(a);    read1(b);}inline void read3(ll &a,ll &b,ll &c){    read1(a);    read1(b);    read1(c);}inline void read1(double &num) {    char in;    double Dec=0.1;    bool IsN=false,IsD=false;    in=getchar();    while(in!='-'&&in!='.'&&(in<'0'||in>'9'))        in=getchar();    if(in=='-') {        IsN=true;        num=0;    } else if(in=='.') {        IsD=true;        num=0;    } else num=in-'0';    if(!IsD) {        while(in=getchar(),in>='0'&&in<='9') {            num*=10;            num+=in-'0';        }    }    if(in!='.') {        if(IsN) num=-num;        return ;    } else {        while(in=getchar(),in>='0'&&in<='9') {            num+=Dec*(in-'0');            Dec*=0.1;        }    }    if(IsN) num=-num;}inline void read2(double &a,double &b){    read1(a);    read1(b);}inline void read3(double &a,double &b,double &c){    read1(a);    read1(b);    read1(c);}inline void hdu(void){    srand(time(0));    printf("%d\n",rand()%5217+1000);}///------------------------------------------------------------------------------const int maxm = 4e4+100;const int mod = 1000003;int pow2[maxm];void init(void){    pow2[0] = 1;    rep0(i,1,maxm) pow2[i] = (pow2[i-1]<<1)%mod; }int gcd(int a,int b,int &x,int &y){    if (b==0){        x=1,y=0;        return a;    }    int q=gcd(b,a%b,y,x);    y-=a/b*x;    return q;}int main(void){//    freopen("in.txt","r",stdin);//    hdu();    int T;    init();    read1(T);    rep1(nouse,1,T)    {        printf("Case %d:",nouse);        int len,x,y;        read1(len);        ll ans = 0;        rep0(i,0,len)         {            read1(x);            ans = (ans+pow2[x])%mod;        }        gcd(len,mod,x,y);        if(x<0) x+=mod;        ans = (ans*x)%mod;        printf("%lld\n",ans);    }    return 0;}