C. Array Game

合集 - 思维题(99)

1.大雪球2024-01-162.E. Final Countdown2024-02-183.D. Divisible Pairs2024-02-194.F. Chat Screenshots2024-02-195.G. One-Dimensional Puzzle2024-02-206.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-217.Watering an Array2024-02-238.Dora and Search2024-02-249.P1137 旅行计划2024-02-2410.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2611.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2612.A. Brick Wall2024-02-2813.B. Minimize Inversions2024-02-2814.C. XOR-distance2024-02-2815.A. Moving Chips2024-02-2816.B. Monsters Attack!2024-02-2817.C. Find B2024-02-2818.D. Slimes2024-02-2819.C. Turtle Fingers: Count the Values of k2024-02-2820.D. Turtle Tenacity: Continual Mods2024-02-2821.D. Vlad and Division2024-02-2822.C. LR-remainders2024-02-2823.P5322 [BJOI2019] 排兵布阵2024-03-0324.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0325.P1156 垃圾陷阱2024-03-0326.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0427.P3047 [USACO12FEB] Nearby Cows G2024-03-0428.P5020 [NOIP2018 提高组] 货币系统2024-03-0429.P3957 [NOIP2017 普及组] 跳房子2024-03-0630.P1550 [USACO08OCT] Watering Hole G2024-03-0631.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0732.P1379 八数码难题2024-03-0733.P6121 [USACO16OPEN] Closing the Farm G2024-03-0834.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0835.P9632 [ICPC2020 Nanjing R] K Co-prime Permutation2024-03-0836.P9825 [ICPC2020 Shanghai R] Fibonacci2024-03-0837.P2034 选择数字2024-03-0938.P7072 [CSP-J2020] 直播获奖2024-03-0939.P1714 切蛋糕2024-03-0940.P4147 玉蟾宫2024-03-1041.P2866 [USACO06NOV] Bad Hair Day S2024-03-1042.P2032 扫描2024-03-1043.P1884 [USACO12FEB] Overplanting S2024-03-1044.D. Lonely Mountain Dungeons2024-03-1245.F. Microcycle2024-03-1246.E. Increasing Subsequences__22024-03-1347.D. Valiant's New Map2024-03-1448.Array Repetition2024-03-1449.F. Rudolf and Imbalance2024-03-1450.D. Exam in MAC2024-03-1551.G. Rudolf and Subway2024-03-1552.C. Ehab and Path-etic MEXs2024-03-1753.A. String Transformation 12024-03-1754.D. Solve The Maze2024-03-1855.B. Array Fix2024-03-1856.B. Equal XOR2024-03-1857.F. Maximum White Subtree2024-03-1958.P1652 圆2024-03-2059.P2181 对角线2024-03-2060.P1325 雷达安装2024-03-2061.P3194 [HNOI2008] 水平可见直线2024-03-2062.B - Make Many Triangles2024-03-2363.[ARC172A] Chocolate2024-03-2564.C. Theofanis' Nightmare2024-03-2765.D. Birthday Gift2024-03-2766.C. Arrow Path2024-03-2767.P1265 公路修建2024-03-2868.P1284 三角形牧场2024-03-2869.Kirill and Mushrooms2024-03-2870.C. Sasha and the Casino2024-03-2871.P1470 [USACO2.3] 最长前缀 Longest Prefix2024-04-0172.P5536 【XR-3】核心城市2024-04-0273.P5836 [USACO19DEC] Milk Visits S2024-04-0274.P4391 [BOI2009] Radio Transmission 无线传输2024-04-0475.P3435 [POI2006] OKR-Periods of Words2024-04-0476.P5960 【模板】差分约束2024-04-0477.P3052 [USACO12MAR] Cows in a Skyscraper G2024-04-0578.E. Long Inversions2024-04-0979.C. Deep Down Below2024-04-0980.C. Rings2024-04-0981.F. Unfair Game2024-04-1082.G. GCD on a grid2024-04-1083.C. Inversion Graph2024-04-1084.Increase Subarray Sums2024-04-1085.D. Cross Coloring2024-04-1286.M. Triangle Construction2024-04-1287.D2. Set To Max (Hard Version)2024-04-12

88.C. Array Game2024-04-12

89.F. Alex's whims2024-04-1290.Queue Sort2024-04-1291.P1908 逆序对2024-04-1392.P5094 [USACO04OPEN] MooFest G 加强版2024-04-1593.P5149 会议座位2024-04-1594.P10337 [UESTCPC 2024] 操作序列2024-04-1595.P10330 [UESTCPC 2024] 黑白珠串2024-04-1596.P10252 线性变换2024-04-1697.P3901 数列找不同2024-04-1698.P10288 [GESP样题 八级] 区间2024-04-1699.P5677 [GZOI2017] 配对统计2024-04-17

收起

原题链接

题解

1.突然遇到新颖的题，我们可以采用逐步变难法、最简策略法、观察数据法
逐步变难法：
当 $k=0$ 时， $an{s}_0=min(a[i])$
$k=1$ 时 $an{s}_1=min(an{s}_0,a[i]-a[i+1](sort))$
观察数据法：
$k=2$ 时 观察数据可知 $n^2$ 左右的算法是可行的，所以我们可以先两端循环遍历所有的第一次 $|a[i]-a[j]|$ ，再二分查找出最接近的值
易错提醒！！：最小值是数组中的最小值而不是相减后的最小值所以还要有一步 $an{s}_2=min(an{s}_2,an{s}_1,an{s}_0)$
最简策略法：
当 $k=3$ ,可以前两次得到相同值，第三次得到0，然后无限循环

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll a[2005]={0};
int main()
{
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,k;
        cin>>n>>k;
        for(ll i=1;i<=n;i++) cin>>a[i];
        sort(a+1,a+n+1,greater<ll>());

        if(k>=3)puts("0");
        else if(k==1)
        {
            ll ans=2e18;
            for(ll i=1;i<n;i++) ans=min(ans,a[i]-a[i+1]);
            for(ll i=1;i<=n;i++) ans=min(ans,a[i]);
            cout<<ans<<endl;
        }
        else
        {
            ll ans=2e18;
            for(ll i=1;i<n;i++) ans=min(ans,a[i]-a[i+1]);
            for(ll i=1;i<=n;i++) ans=min(ans,a[i]);
            for(ll i=1;i<=n-1;i++)
            {
                for(ll j=i+1;j<=n;j++)
                {
                    ll l=1,r=n+1;
                    ll d1=a[i]-a[j];
                    while(l+1<r)
                    {
                        ll mid=(l+r)/2;
                        if(a[mid]>=d1) l=mid;
                        else r=mid;
                    }
                    if(l!=n) ans=min(min(abs(d1-a[r]),abs(d1-a[l])),ans);
                    else ans=min(ans,abs(d1-a[l]));
                }
            }
            cout<<ans<<endl;
        }
    }
    return 0;
}