TYVJ 1020 寻找质因子

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描述 Description

数轴上有N个点，任意两点连线得到n(n-1)条线段，试求线段的总长。

输入格式 InputFormat

第一行，一个整数N，表示点数。
接下来N行，每行一个整数X_i，表示点的坐标。

输出格式 OutputFormat

一个整数，表示线段的总长。

样例输入 SampleInput [复制数据]

515324

样例输出 SampleOutput [复制数据]

40

数据范围和注释 Hint

N <= 10000 , 0 <= X_i <= 1000000000

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <map>#include <cmath>#include <string>#include <queue>#include <set>using namespace std;#ifdef WINtypedef __int64 LL;#define form "%I64d\n"#elsetypedef long long LL;#define form "%lld\n"#endif#define SI(a) scanf("%d", &(a))#define SDI(a, b) scanf("%d%d", &(a), &(b))#define S64I(a) scanf(form, &a)#define SS(a) scanf("%s", (a))#define SDS(a, b) scanf("%s%s", (a), (b))#define SC(a) scanf("%c", &(a))#define PI(a) printf("%d\n", a)#define PS(a) puts(a)#define P64I(a) printf(form, a)#define Max(a, b) (a > b ? a : b)#define Min(a, b) (a < b ? a : b)#define MSET(a, b) (memset(a, b, sizeof(a)))#define Mid(L, R) (L + (R - L)/2)#define Abs(a) (a > 0 ? a : -a)#define REP(i, n) for(int i=0; i < (n); i++)#define FOR(i, a, n) for(int i=(a); i <= (n); i++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const int maxa = 20000 + 50;int vis[maxa], prime[maxa], pcnt;void sieve(int n) {int m = (int) sqrt(n + 0.5);MSET(vis, 0);for(int i=2; i<=n; i++) if(!vis[i]) {for(int j=i*i; j<=n; j+=i) vis[j] = 1;}}int get_prime(int n) {sieve(n);int cnt = 0;for(int i=2; i<=n; i++) if(!vis[i]) {prime[cnt++] = i;}return cnt;}int de(int n) {int res = 1;int t = n;for(int i=0; i<pcnt; i++) {while(t % prime[i] == 0) {res = max(res, prime[i]);t /= prime[i];}if(!vis[t]) {res = max(res, t);break;}}return res;}int main() {int n, mm = 0, ans;pcnt = get_prime(maxa);SI(n);while(n--) {int a;SI(a);int t = de(a);if(t > mm) {mm = t;ans = a;}}PI(ans);return 0;}