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  • CodeForces 731F Video Cards (数论+暴力)

    题意:给定 n 个数,可以对所有的数进行缩小,问你找出和最大的数,使得这些数都能整除这些数中最小的那个数。

    析:用前缀和来做,先统计前 i 个数中有有多少数,然后再进行暴力去找最大值,每次都遍历这一段就好。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #define debug puts("+++++")
    //#include <tr1/unordered_map>
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    //using namespace std :: tr1;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const LL LNF = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 2e5 + 5;
    const LL mod = 1e9 + 7;
    const int N = 1e6 + 5;
    const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
    const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
    const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
    inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
    inline int lcm(int a, int b){  return a * b / gcd(a, b); }
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline int Min(int a, int b){ return a < b ? a : b; }
    inline int Max(int a, int b){ return a > b ? a : b; }
    inline LL Min(LL a, LL b){ return a < b ? a : b; }
    inline LL Max(LL a, LL b){ return a > b ? a : b; }
    inline bool is_in(int r, int c){
        return r >= 0 && r < n && c >= 0 && c < m;
    }
    int a[maxn];
    int sum[maxn];
    
    int main(){
        while(scanf("%d", &n) == 1){
            int x;
            memset(a, 0, sizeof a);
            for(int i = 0; i < n; ++i){
                scanf("%d", &x);
                ++a[x];
            }
            for(int i = 1; i <= 200000; ++i)  sum[i] = sum[i-1] + a[i];
            LL ans = 0;
            for(int i = 1; i <= 200000; ++i) if(a[i]){
                LL tmp = 0;
                int j;
                for(j = i-1; j <= 200000; j += i)
                    tmp += (LL)(sum[j] - sum[j-i]) * (j+1-i);
                if(j > 200000)  tmp += (LL)(sum[200000]-sum[j-i]) * (j+1-i);
                ans = Max(ans, tmp);
            }
            cout << ans << endl;
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5971589.html
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