POJ1995 Raising Modulo Numbers（乘方取余）

Raising Modulo Numbers

http://poj.org/problem?id=1995

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4434 Accepted: 2544

Description

People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow: 

Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai^Bi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers. 

You should write a program that calculates the result and is able to find out who won the game. 

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression 

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

Sample Input

31642 33 44 55 63612312374859 30293821713 18132

Sample Output

21319513

 

方法一：

将b以二进制存在数组中，再利用公式：

a^b=a^{2^b0+2^b1+…+2^bk}=a^{2^b0}×a^{2^b1}×a^{2^bk}

其中因子递推，后项因子为前项的平方；

本题关键在于：

为防止中间结果过大溢出，应不断取模

 

#include<stdio.h>long a[45004],b[45004];int idx[34],am[34];int abm(long a,long b,int m){    int i=0,j,k,p=1;    while(b)    {        idx[i++]=b%2;        b>>=1;    }    am[0]=a%m;    for(j=1; j<i; j++)        am[j]=am[j-1]*am[j-1]%m;    for(k=0; k<i; k++)    {        if(idx[k])            p=p*am[k]%m;    }    return p;}int main(){    int z,m,h,i,s;    scanf("%d",&z);    while(z--)    {        scanf("%d%d",&m,&h);        for(i=0; i<h; i++)            scanf("%ld%ld",&a[i],&b[i]);        s=0;        for(i=0; i<h; i++)            s=(s+abm(a[i],b[i],m))%m;        printf("%d\n",s);    }    return 0;}

 

 方法二：

#include<stdio.h>long long a[45004],b[45004];long long quick_mod(long long a,long long b,long long m){    long long ans=1;    a%=m;    while(b)    {        if(b&1)        {            ans=(ans*a)%m;            b--;        }        b>>=1;        a=(a*a)%m;    }    return ans;}int main(){    int z,m,h,i,s;    scanf("%d",&z);    while(z--)    {        scanf("%d%d",&m,&h);        for(i=0; i<h; i++)            scanf("%lld%lld",&a[i],&b[i]);        s=0;        for(i=0; i<h; i++)            s=(s+quick_mod(a[i],b[i],m))%m;        printf("%d\n",s);    }    return 0;}

附：乘方取余模板

long long multi(long long a,long long b,long long m)//a*b%m{    long long ret=0;    while(b>0)    {        if(b&1)ret=(ret+a)%m;        b>>=1;        a=(a<<1)%m;    }    return ret;}long long quick_mod(long long a,long long b,long long m)//a^b%m{    long long ans=1;    a%=m;    while(b)    {        if(b&1)        {            ans=multi(ans,a,m);//或者写成ans=(ans*a)%m            b--;        }        b>>=1;        a=multi(a,a,m);//或者写成a=(a*a)%m    }    return ans;}

Raising Modulo Numbers

http://poj.org/problem?id=1995

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4434 Accepted: 2544

Description

People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow: 

Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai^Bi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers. 

You should write a program that calculates the result and is able to find out who won the game. 

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression 

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

Sample Input

31642 33 44 55 63612312374859 30293821713 18132

Sample Output

21319513

 

方法一：

将b以二进制存在数组中，再利用公式：

a^b=a^{2^b0+2^b1+…+2^bk}=a^{2^b0}×a^{2^b1}×a^{2^bk}

其中因子递推，后项因子为前项的平方；

本题关键在于：

为防止中间结果过大溢出，应不断取模

 

#include<stdio.h>long a[45004],b[45004];int idx[34],am[34];int abm(long a,long b,int m){    int i=0,j,k,p=1;    while(b)    {        idx[i++]=b%2;        b>>=1;    }    am[0]=a%m;    for(j=1; j<i; j++)        am[j]=am[j-1]*am[j-1]%m;    for(k=0; k<i; k++)    {        if(idx[k])            p=p*am[k]%m;    }    return p;}int main(){    int z,m,h,i,s;    scanf("%d",&z);    while(z--)    {        scanf("%d%d",&m,&h);        for(i=0; i<h; i++)            scanf("%ld%ld",&a[i],&b[i]);        s=0;        for(i=0; i<h; i++)            s=(s+abm(a[i],b[i],m))%m;        printf("%d\n",s);    }    return 0;}

 

 方法二：

#include<stdio.h>long long a[45004],b[45004];long long quick_mod(long long a,long long b,long long m){    long long ans=1;    a%=m;    while(b)    {        if(b&1)        {            ans=(ans*a)%m;            b--;        }        b>>=1;        a=(a*a)%m;    }    return ans;}int main(){    int z,m,h,i,s;    scanf("%d",&z);    while(z--)    {        scanf("%d%d",&m,&h);        for(i=0; i<h; i++)            scanf("%lld%lld",&a[i],&b[i]);        s=0;        for(i=0; i<h; i++)            s=(s+quick_mod(a[i],b[i],m))%m;        printf("%d\n",s);    }    return 0;}

附：乘方取余模板

long long multi(long long a,long long b,long long m)//a*b%m{    long long ret=0;    while(b>0)    {        if(b&1)ret=(ret+a)%m;        b>>=1;        a=(a<<1)%m;    }    return ret;}long long quick_mod(long long a,long long b,long long m)//a^b%m{    long long ans=1;    a%=m;    while(b)    {        if(b&1)        {            ans=multi(ans,a,m);//或者写成ans=(ans*a)%m            b--;        }        b>>=1;        a=multi(a,a,m);//或者写成a=(a*a)%m    }    return ans;}