poj1995 Raising Modulo Numbers【高速幂】
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 5500 | Accepted: 3185 |
Description
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 AiBi 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
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
(A1B1+A2B2+ ... +AHBH)mod M.
Sample Input
3
16
4
2 3
3 4
4 5
5 6
36123
1
2374859 3029382
17
1
3 18132
Sample Output
2
13195
13
#include<stdio.h>
#include<string.h>
int ksm(long long a,long long b,long long n)
{
long long ans=1;
while(b)
{
if(b&1)
ans=ans*a%n;
//写成ans*=a%n不行...输出错误
a=a*a%n;
b>>=1;
}
return ans;
}
int main()
{
int z,h,m,u,i;
scanf("%d",&z);
while(z--)
{
int sum = 0;
scanf("%d",&m);
scanf("%d",&h);
while(h--)
{
scanf("%d%d",&u,&i);
sum += ksm(u,i,m);
sum %= m;
}
printf("%d\n",sum);
}
return 0;
} /*************************************************************/ #include<stdio.h>
#include<string.h>
int main()
{
long long z,h,m,a,b,ans;
scanf("%lld",&z);
while(z--)
{
int sum = 0;
scanf("%lld",&m);
scanf("%lld",&h);
while(h--)
{
scanf("%lld%lld",&a,&b);
ans = 1;
while(b)
{
if(b&1)
ans=ans*a%m;
a=a*a%m;
b>>=1;
}
sum += ans;
sum %= m;
}
printf("%d\n",sum%m);
}
return 0;
}
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