心路歷程

預計得分：\(100 + 100 + 100 = 300\)

實際得分：\(100 +100 +100 = 300\)

學OI兩年終於AK了一次qwq(雖然題目炒雞水。。)

紀念一下這令人激動的時刻。。

8點開始考，9:40就都拍上了。。可見題目確實水。。然後又去做了做另一套

Sol

T1

題目中給的兩個陣列沒啥卵用，都是可以直接求的

然後離散化+樹狀陣列就完了

#include<cstdio>
#include<algorithm>
#include<iostream>
#define lb(x) (x & (-x))
#define LL long long 
using namespace std;
const LL MAXN = 1e6 + 10;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
LL N, f[MAXN], g[MAXN], a[MAXN], mod;
LL add(LL x, LL y) {
    if(x + y < 0) return x + y + mod;
    return x + y >= mod ? x + y - mod : x + y;
}
LL mul(LL x, LL y) {
    return 1ll * x * y % mod;
}
LL fp(LL a, LL p) {
    LL base = 1;
    while(p) {
        if(p & 1) base = mul(base, a); 
        a = mul(a, a); p >>= 1;
    }
    return base;
}
LL date[MAXN], num;
void Des() {
    sort(date + 1, date + num + 1);
    num = unique(date + 1, date + num + 1) - date - 1;
    for(LL i = 1; i <= N; i++) f[i] = lower_bound(date + 1, date + num + 1, f[i]) - date, 
                                g[i] = lower_bound(date + 1, date + num + 1, g[i]) - date;
}
LL T[MAXN];
void Add(LL x, LL val) {
    while(x <= num) T[x] += val, x += lb(x);
}
LL Query(LL x) {
    LL ans = 0;
    while(x) ans += T[x], x -= lb(x);
    return ans;
}
int main() { 
    freopen("calc.in", "r", stdin);
    freopen("calc.out", "w", stdout);
    N = read(); mod = read();
    for(LL i = 1; i <= N; i++) a[i] = read(), f[i] = fp(i, a[i]), g[i] = fp(a[i], i), 
                                               date[++num] = f[i], date[++num] = g[i];
    Des();
    LL ans = 0;
    for(LL i = N; i >= 1; i--) {
        ans += Query(f[i] - 1);
        Add(g[i], 1);
    }
    cout << ans;
    return 0;
}
 

T2

這題比較interesting啊。

我是先做的T3再回來做的T2

首先二分答案

很顯然的一個貪心是從左往右掃，如果遇到一個不合法的點\(i\)，那麼升級\(i + R\)處的炮臺。。

和暴力拍了10000多組沒錯誤。。贏了

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define LL long long 
using namespace std;
const LL MAXN = 2e6;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
LL N, R;
LL  tag[MAXN], s[MAXN], a[MAXN], b[MAXN], K;
bool check(LL val) {
    memset(tag, 0, sizeof(tag)); 
    LL now = 0, res = K, flag = 1;
    for(LL i = 1; i <= N; i++) {
        now += tag[i];
        a[i] += now;
        if(a[i] < val) now += val - a[i], tag[i + 2 * R + 1] -= val - a[i], res -= val - a[i];
        if(res < 0) {flag = 0; break;}
    }
    memcpy(a, b, sizeof(a));
    return flag;
}
LL Query(LL l, LL r) {
    r = min(r, N);
    if(l < 0) return s[r];
    return s[r] - s[l];
}
int main() {
    freopen("game.in", "r", stdin); freopen("game.out", "w", stdout);
    N = read(); R = read(); K = read();
    for(LL i = 1; i <= N; i++) a[i] = read(), s[i] = s[i - 1] + a[i]; 
    for(LL i = 1; i <= N; i++) a[i] = Query(i - R - 1, i + R);
    for(LL i = 1; i <= N; i++) b[i] = a[i];
    LL l = 0, r = 3e18, ans = 0;
    while(l <= r) {
        LL mid = (l + r) >> 1;
        if(check(mid)) l = mid + 1, ans = mid;
        else r = mid - 1;
    }
    cout << ans;
    return 0;
}
 

T3

出題人還能再套路一點麼。。

首先把詢問離線後從小到大排序

合併的時候用帶權並查集維護一下

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#define Pair pair<LL, LL>
#define MP make_pair
#define fi first
#define se second 
#define LL long long 
using namespace std;
const LL MAXN = 1e6 + 10, INF = 1e9 + 10;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
LL N, M, Q, cnt, fa[MAXN], w[MAXN], ans, out[MAXN];
struct Edge {
    LL u, v, w;
    bool operator < (const Edge &rhs) const {
        return w < rhs.w;
    }
}E[MAXN];
Pair q[MAXN];
LL find(LL x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
void AddEdge(LL i) {
    LL x = E[i].u, y = E[i].v, v = E[i].w;
    LL fx = find(x), fy = find(y);
    if(fx == fy) w[fx] += v, ans = max(ans, w[fx]);
    else {
        fa[fx] = fy; w[fy] += v + w[fx]; w[fx] = 0;
        ans = max(ans, w[fy]);
    }
}
int main() {
    freopen("graph.in", "r", stdin); freopen("graph.out", "w", stdout);
    N = read(); M = read(); Q = read();
    for(LL i = 1; i <= N; i++) fa[i] = i;
    for(LL i = 1; i <= M; i++) E[i].u = read(), E[i].v = read(), E[i].w = read();
    sort(E + 1, E + M + 1);
    for(LL i = 1; i <= Q; i++) q[i] = MP(read(), i);
    sort(q + 1, q + Q + 1);
    LL j = 1;
    for(LL i = 1; i <= Q; i++) {
        while(j <= M && E[j].w <= q[i].fi) 
            AddEdge(j++);
        out[q[i].se] = ans;
    }
    for(LL i = 1; i <= Q; i++) printf("%I64d\n", out[i]);
    return 0;
}