P1198 [JSOI2008] 最大数

合集 - 洛谷(99)

1.P8755 [蓝桥杯 2021 省 AB2] 负载均衡2023-11-222.P1102 A-B 数对的三种解法2023-11-263.P1090 [NOIP2004 提高组] 合并果子2023-11-264.P3379 【模板】最近公共祖先（LCA）2023-11-285.P8599 [蓝桥杯 2013 省 B] 带分数2023-11-286.P1433 吃奶酪2023-11-297.P3385 【模板】负环2023-12-068.P1576 最小花费2023-12-069.Zuma2023-12-0610.P3205 [HNOI2010] 合唱队2023-12-0611.P1220 关路灯2023-12-0712.P4170 [CQOI2007] 涂色（天赋哥不要点进来）2023-12-1113.P8805 [蓝桥杯 2022 国 B] 机房2023-12-1214.P3386 【模板】二分图最大匹配2023-12-2015.P2197 【模板】Nim 游戏2023-12-2016.P1036 [NOIP2002 普及组] 选数2023-12-2117.T397291 【模板】拓扑排序（加强版）2023-12-2618.P2865 [USACO06NOV] Roadblocks G2023-12-2619.P1168 中位数2024-01-0220.P1631 序列合并2024-01-0221.P3372 【模板】线段树 12024-01-0222.P1352 没有上司的舞会2024-01-0223.P2015 二叉苹果树2024-01-0324.P2014 [CTSC1997] 选课2024-01-0325.P1613 跑路2024-01-0326.P2911 [USACO08OCT] Bovine Bones G2024-01-0427.P4316 绿豆蛙的归宿2024-01-0428.P3870 [TJOI2009] 开关2024-01-1029.P3373 【模板】线段树 22024-01-1430.P1558 色板游戏2024-01-1531.P1638 逛画展2024-01-1632.P1043 [NOIP2003 普及组] 数字游戏2024-01-2633.P1147 连续自然数和2024-01-2634.P1063 [NOIP2006 提高组] 能量项链2024-01-2735.P4342 [IOI1998] Polygon2024-01-2736.P1122 最大子树和2024-02-0537.P2016 战略游戏2024-02-0538.P2585 [ZJOI2006] 三色二叉树2024-02-0739.P1273 有线电视网2024-02-0740.P2986 [USACO10MAR] Great Cow Gathering G2024-02-0741.P1114 “非常男女”计划2024-02-0842.P5026 Lycanthropy2024-02-0843.P4231 三步必杀2024-02-0944.P2985 [USACO10FEB] Chocolate Eating S2024-02-0945.P4090 [USACO17DEC] Greedy Gift Takers P2024-02-1046.P4933 大师2024-02-1047.P1082 [NOIP2012 提高组] 同余方程2024-02-1148.P3811 【模板】模意义下的乘法逆元2024-02-1249.P9725 [EC Final 2022] Chase Game 22024-02-1250.P1330 封锁阳光大学2024-02-1251.P8674 [蓝桥杯 2018 国 B] 调手表2024-02-1252.P10111 [GESP202312 七级] 纸牌游戏2024-02-1453.P9975 [USACO23DEC] Cowntact Tracing 2 B2024-02-1554.P8783 [蓝桥杯 2022 省 B] 统计子矩阵2024-02-1555.P8786 [蓝桥杯 2022 省 B] 李白打酒加强版2024-02-1556.P8807 [蓝桥杯 2022 国 C] 取模2024-02-1557.P8732 [蓝桥杯 2020 国 ABC] 答疑2024-02-1558.P8725 [蓝桥杯 2020 省 AB3] 画中漂流2024-02-1559.P3958 [NOIP2017 提高组] 奶酪2024-02-1560.P4084 [USACO17DEC] Barn Painting G2024-02-1561.P1892 [BOI2003] 团伙2024-02-16

62.P1198 [JSOI2008] 最大数2024-02-16

63.P1204 [USACO1.2] 挤牛奶Milking Cows2024-02-1664.P5490 【模板】扫描线2024-02-1765.P8784 [蓝桥杯 2022 省 B] 积木画2024-02-1766.P9325 [CCC 2023 S2] Symmetric Mountains2024-02-1767.Iva & Pav2024-02-1768.P9889 [ICPC2018 Qingdao R] Plants vs. Zombies2024-02-1769.P9847 [ICPC2021 Nanjing R] Crystalfly2024-02-1870.E. Anna and the Valentine's Day Gift2024-02-1971.F. Chat Screenshots2024-02-1972.G. One-Dimensional Puzzle2024-02-2073.P2899 [USACO08JAN] Cell Phone Network G2024-02-2074.P2023 [AHOI2009] 维护序列2024-02-2075.P1531 I Hate It2024-02-2076.P5057 [CQOI2006] 简单题2024-02-2077.P2357 守墓人2024-02-2078.P8775 [蓝桥杯 2022 省 A] 青蛙过河2024-02-2179.P1656 炸铁路2024-02-2280.Watering an Array2024-02-2381.Dora and Search2024-02-2482.P1137 旅行计划2024-02-2483.P2835 刻录光盘2024-02-2484.P1197 [JSOI2008] 星球大战2024-02-2485.P3388 【模板】割点（割顶）2024-02-2486.P8435 【模板】点双连通分量2024-02-2587.P8436 【模板】边双连通分量2024-02-2588.P2860 [USACO06JAN] Redundant Paths G2024-02-2589.P3067 [USACO12OPEN] Balanced Cow Subsets G2024-02-2690.P4799 [CEOI2015 Day2] 世界冰球锦标赛2024-02-2691.P6492 [COCI2010-2011#6] STEP2024-02-2692.P1653 [USACO04DEC] Cow Ski Area G2024-02-2693.P1040 [NOIP2003 提高组] 加分二叉树2024-03-0394.P5322 [BJOI2019] 排兵布阵2024-03-0395.P2946 [USACO09MAR] Cow Frisbee Team S2024-03-0396.P1156 垃圾陷阱2024-03-0397.P1064 [NOIP2006 提高组] 金明的预算方案2024-03-0498.P3047 [USACO12FEB] Nearby Cows G2024-03-0499.P2466 [SDOI2008] Sue 的小球2024-03-04

收起

原题链接

题解1：单调栈+并查集？

单调栈特性：栈内元素大小和序号由栈底到栈顶具有单调性，本题大小单调减，序号单调增
维护：新元素入栈时，栈内剩余的所有小于该元素的元素出栈，并视新元素为集合首领，然后新元素入栈
查询：查询集合首领即可

code1

#define ll long long
#include<bits/stdc++.h>
using namespace std;

struct num {
    ll pos;
    ll val;
};

ll fa[200005] = {0};

ll finds(ll now) {
    return fa[now] = (now == fa[now] ? now : finds(fa[now]));
}

vector<ll> a;

inline void read(ll &x) {  
    x = 0;
    ll flag = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') flag = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48); 
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x) {
    if(x < 0) {
        putchar('-');
        x = -x;
    }
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

int main() {
    ll m, d;
    read(m); read(d);

    for(ll i = 0; i < m; i++) fa[i] = i;

    ll t = 0, len = 0;
    stack<num> q;
    while(m--) {
        char op;
        ll n;
        scanf(" %c", &op);
        read(n);
        if(op == 'Q') {
            n = a.size() - n ;
            write(a[finds(n)]);
            putchar('\n');
            t = a[finds(n)];
        } else {
            a .push_back (n = (n + t) % d);
            while(!q.empty() && n >= q.top().val) {
                ll now = q.top().pos;
                fa[finds(now)] = a.size()-1;
                q.pop();
            }
            q.push({a.size()-1, n});
        }
    }
    return 0;
}

题解2

st表，不过是向左延申的st表

code

#define ll long long
#include<bits/stdc++.h>
using namespace std;

// Custom input and output functions
inline void read(ll &x) {  
    x = 0;
    ll flag = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') flag = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9') {
        x = (x << 3) + (x << 1) + (c ^ 48); 
        c = getchar();
    }
    x *= flag;
}

inline void write(ll x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

ll len = 0;
vector<vector<ll>> f(200005, vector<ll>(35, 0)); // Using vector for f
ll a[200005] = {0}; // Keeping a as an array since it's directly accessed by index and there's no mention of dynamic resizing

void change() {
    f[len][0] = a[len];
    for (ll i = 1; (1LL << i) <= len; i++) {
        f[len][i] = max(f[len][i - 1], f[len - (1LL << (i - 1))][i - 1]);
    }
}

int main() {
    ll n, d;
    read(n); // Using custom read function
    read(d); // Using custom read function
    ll t = 0;
    while (n--) {
        char op;
        cin >> op; // Keeping cin for char as it's simple and efficient
        ll x;
        read(x); // Using custom read function
        if (op == 'A') {
            a[++len] = (x + t) % d;
            change();
        } else {
            ll ans = 0;
            ll l = len - x + 1; // Ensuring correct value of l
            ll r = len;
            for (ll j = 31; j >= 0; j--) {
                if ((1LL << j) <= r - l + 1) { // Using 1LL to ensure long long calculation
                    ans = max(ans, f[r][j]);
                    r -= 1LL << j; // Adjusting r for the new position
                }
            }

            write(ans); // Using custom write function
            putchar('\n'); // To end the line after output
            t = ans;
        }
    }
    return 0;
}