upc组队赛17 Greatest Common Divisor【gcd+最小质因数】

Greatest Common Divisor

题目链接

题目描述

There is an array of length n, containing only positive numbers.
Now you can add all numbers by 1 many times. Please find out the minimum times you need to perform to obtain an array whose greatest common divisor(gcd) is larger than 1 or state that it is impossible.
You should notice that if you want to add one number by 1, you need to add all numbers by 1 at the same time.

输入

The first line of input file contains an integer T (1≤T≤20), describing the number of test cases.
Then there are 2×T lines, with every two lines representing a test case.
The first line of each case contains a single integer n (1≤n≤105) described above.
The second line of that contains n integers ranging in [1,109].

输出

Please output T lines exactly.
For each line, output Case d: (d represents the order of the test case) first. Then output the answer in the same line. If there is no way for that, print -1 instead.

样例输入

3
1
2
5
2 5 9 5 7
5
3 5 7 9 11

样例输出

Case 1: 0
Case 2: -1
Case 3: 1

提示

Sample 1: You do not need to do anything because its gcd is already larger than 1.
Sample 2: It is impossible to obtain that array.
Sample 3: You just need to add all number by 1 so that gcd of this array is 2.

题意

问几次操作能使所有数的最小公约数不为1

题解

首先需要求出各个数之间的差，求出各个差的gcd；
然后找到gcd的最小质因数g ，操作的次数为距离a[0]最近的g的倍数-a[0]； 即g - a[0] % g;
还有其他各种特判，在代码中说明

代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define scac(x) scanf("%c",&x)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d
",x)
#define pri2(x,y) printf("%d %d
",x,y)
#define pri3(x,y,z) printf("%d %d %d
",x,y,z)
#define prl(x) printf("%lld
",x)
#define prl2(x,y) printf("%lld %lld
",x,y)
#define prl3(x,y,z) printf("%lld %lld %lld
",x,y,z)
#define mst(x,y) memset(x,y,sizeof(x))
#define ll long long
#define LL long long
#define pb push_back
#define mp make_pair
#define P pair<double,double>
#define PLL pair<ll,ll>
#define PI acos(1.0)
#define eps 1e-6
#define inf 1e17
#define mod 1e9+7
#define INF 0x3f3f3f3f
#define N 1005
const int maxn = 1e5+10;
ll a[maxn],b[maxn];
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
int n;
int  main()
{
    int t;
    sca(t);
    int cas = 1;
    while(t--)
    {
        sca(n);
        rep(i,0,n)  scl(a[i]);
        if(n == 1) //如果只有一个数
        {
            if(a[0] == 1) printf("Case %d: 1
",cas++);
            else  printf("Case %d: 0
",cas++);
            continue;
        }
        sort(a,a+n);
        int cnt = 0;
        rep(i,1,n)
        {
            if(a[i] != a[i-1]) //如果数相同就跳过
                b[cnt++] = a[i] - a[i-1];
        }
        if(cnt == 0)//如果所有数都相同
        {
            if(a[0] == 1) printf("Case %d: 1
",cas++);
            else  printf("Case %d: 0
",cas++);
            continue;
        }
        ll g = b[0];
        rep(i,1,cnt)  g = gcd(g,b[i]);
        if(g == 1) //不可能的情况
        {
            printf("Case %d: -1
",cas++);
            continue;
        }
        if(gcd(g,a[0]) > 1) g = gcd(g,a[0]);
        rep(i,2,100000)
        {
          if(g % i == 0)
          {
            g = i;
            break;
          }
        }
        ll ans;
        if(a[0] % g) ans = g - a[0] % g;
        else ans = 0;
        printf("Case %d: %lld
",cas++,ans);
    }
}