HDOJ 2824 The Euler function(欧拉函数+打表法)

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The Euler function

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4516 Accepted Submission(s): 1878

Problem Description

The Euler function phi is an important kind of function in number theory, (n) represents the amount of the numbers which are smaller than n and coprime to n, and this function has a lot of beautiful characteristics. Here comes a very easy question: suppose you are given a, b, try to calculate (a)+ (a+1)+....+ (b)

Input

There are several test cases. Each line has two integers a, b (2<a<b<3000000).

Output

Output the result of (a)+ (a+1)+....+ (b)

Sample Input

3 100

Sample Output

题意：给出a，b两个数。(n)表示小于或者等于n的正整数中与n互质的个数。求(a)+(a+1)+.....+(b)的值。

很明显是欧拉函数问题，a，b值的上限很大，所以必须打表。不懂打表代码，只有死记模板了。

欧拉函数打表，代码如下：

<span style="font-size:12px;">#include<cstdio>#include<cstring>#define maxn 3000010__int64 res[maxn];__int64 euler(){int i,j;memset(res,0,sizeof(res));res[1]=1;for(i=2;i<maxn;++i)//先打表出每个数的互质数个数 {if(!res[i]){for(j=i;j<maxn;j+=i){if(!res[j])res[j]=j;res[j]=res[j]/i*(i-1);}}}for(i=2;i<maxn;++i)//再循环相加，找出从1到这个数的互质数和    res[i]+=res[i-1];}int main(){euler();int a,b;while(scanf("%d%d",&a,&b)!=EOF)   printf("%I64d\n",res[b]-res[a-1]);return 0;}</span>

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