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  • UVALive 6692 Lucky Number (思路 + 枚举)

    题意:给你n 个数字,某一个数的幸运数是这个数前面比他小 离他最远的位置之差,求出最大幸运值。

    析:先按从大到小排序,然后去维护那个最大的id,一直比较,更新最大值就好。

    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 freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 0x3f3f3f3f3f3f;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1e6 + 5;
    const int mod = 1e8;
    const char *mark = "+-*";
    const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
    const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
    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;
    }
    struct node{
        int x, id;
        bool operator < (const node &p) const{
            return x > p.x;
        }
    };
    node a[maxn];
    
    int main(){
        int T;   cin >> T;
        while(T--){
            scanf("%d", &n);
            for(int i = 0; i < n; ++i){
                scanf("%d", &a[i].x);
                a[i].id = i;
            }
            sort(a, a+n);
            int ans = 0, mmax = -1, tmp = -1;
            for(int i = 0; i < n; ++i){
                ans = max(ans, mmax-a[i].id);
                tmp = max(tmp, a[i].id);
                if(i != n-1 && a[i].x > a[i+1].x)
                    mmax = max(tmp, mmax);
            }
            printf("%d
    ", ans);
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5800867.html
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