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http://acm.hdu.edu.cn/showproblem.php?pid=1695

GCD

2
1 3 1 5 1
1 11014 1 14409 9

Case 1: 9
Case 2: 736427

Hint
For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
 

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#define maxn 100000
using namespace std;
int phi[100005],fac[1000];
int a,b,c,d,k;
vector<int> x[maxn+5];
bool is[maxn+5];
void get_prim(){
    memset(is, false, sizeof(is));
    for (int i=0; i<maxn; i++) x[i].clear();
    for (int j=2; j<maxn; j+=2) x[j].push_back(2);
    for (int i=3; i<maxn; i+=2)
        if (!is[i]) {
            for (int j=i; j<maxn; j+=i) {
                is[j] = true;
                x[j].push_back(i);
            }
        }
}
void euler(){ //返回euler(n)
    for (int i = 1; i <= maxn; i++) phi[i] = i;
    for (int i = 2; i <= maxn; i += 2) phi[i] /= 2;
    for (int i = 3; i <= maxn; i += 2)
              if(phi[i] == i) {
        for (int j = i; j <= maxn; j += i)
            phi[j] = phi[j] / i * (i - 1);
          }
}
int get_ans(int z){
    int cnt=x[z].size();
    int sum=0;
    for(int num=1;num<(1<<cnt);num++){//cnt个集合，容斥定理中就有2的cnt次方-1项,num是二进制压缩//3为11，代表A1交A2
        int mult=1,ones=0;
        for(int i=0;i<cnt;i++){//i表示第几个集合，例如i=2，表示为100，表示为第三个集合
            if(num&(1<<i)){//当i为0的时候，判断第一个集合是否进行交集操作
                ones++;//当ones为偶数的时候，那么就是减，当noes为奇数的时候为加
                mult*=x[z][i];
            }
        }
        if(ones%2) sum+=b/mult;
        else sum-=b/mult;
    }
    return b-sum;
}
int main()
{
    //freopen("F://in.txt","r",stdin);
    euler();
    get_prim();
    int t;
    scanf("%d",&t);
    for(int Case=1;Case<=t;Case++){
        __int64 ans=0;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(b>d)swap(b,d);
        if(k==0||k>d){
            printf("Case %d: %I64d\n",Case,ans);
            continue;
        }
        b/=k;
        d/=k;
        for(int i=1;i<=b;i++)//printf("%d\n",phi[i]);
        ans+=phi[i];
        //printf("%I64d\n",ans);
        for(int i=b+1;i<=d;i++)
            ans+=get_ans(i);
            printf("Case %d: %I64d\n",Case,ans);
    }
}

GCD

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原文地址：http://blog.csdn.net/burning_newbie/article/details/48001235