#1：来自辣鸡的初奏——6

joyoi楼兰图腾，树状数组求逆序对的题，joyoi过不去，contesthunter过去了……

#include <cstdio>
#include <cstring>
#define ll long long
#define init(a, b) memset(a, b, sizeof(a))

const int maxn = 200005;
int n;
int a[maxn], t[maxn];
ll l[maxn], r[maxn];

int ask(int x) {
    int ret = 0;
    for (; x; x -= x & (-x))    ret += t[x];
    return ret;
}

void add(int x, int cnt) {
    for (; x <= n; x += x & (-x))
        t[x] += cnt;
}

ll All() {
    ll ret = 0ll;
    for (int i = 0; i < n; i++)
        ret += l[i] * r[i];
    return ret;
}

ll solve0() {
    init(t, 0);
    init(l, 0);
    init(r, 0);
    for (int i = 0; i < n; i++) {
        l[i] += ask(n) - ask(a[i]);
        add(a[i], 1);
    }
    init(t, 0);
    for (int i = n-1; ~i; i--) {
        r[i] += ask(n) - ask(a[i]);
        add(a[i], 1);
    }
    return All();
}

ll solve1() {
    init(t, 0);
    init(l, 0);
    init(r, 0);
    for (int i = 0; i < n; i++) {
        l[i] += ask(a[i]-1);
        add(a[i], 1);
    }
    init(t, 0);
    for (int i = n-1; ~i; i--) {
        r[i] += ask(a[i]-1);
        add(a[i], 1);
    }
    return All();
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &a[i]);
    }
    printf("%lld %lld", solve0(), solve1());
}

 

UVALive4329，枚举标杆是常见手段，其余的跟上一个题一样。然后发现上一个题写得过于直观可以优化。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxa 100001
#define maxn 20005
#define init(a, b) memset(a, b, sizeof(a))
#define ll long long

int n, T, maxx;
int a[maxn], l[maxn], r[maxn];
int t[maxa];

int ask(int x) {
    int ret = 0;
    for (; x; x -= x & -x)  ret += t[x];
    return ret;
}

void add(int x, int cnt) {
    for (; x <= maxx; x += x & -x)  t[x] += cnt;
}

ll solve() {
    init(t, 0);
    init(l, 0);
    init(r, 0);  
    for (int i = 0; i < n; i++) {
        l[i] += ask(maxx) - ask(a[i]); 
        add(a[i], 1);
    }
    init(t, 0);
    for (int i = n-1; ~i; i--) {
        r[i] += ask(a[i] - 1);
        add(a[i], 1);
    }

    ll ret = 0ll;
    for (int i = 1; i < n-1; i++)
        ret += (ll)l[i] * r[i] + (ll)(i-l[i]) * (n-1-i-r[i]);
    return ret;
}

int main() {
    for (scanf("%d", &T); T; T--) {
        maxx = -1;
        scanf("%d", &n);
        for (int i = 0; i < n; i++) {
            scanf("%d", &a[i]);
            maxx = max(maxx, a[i]);
        }
        printf("%lld\n", solve());
    }
    return 0;
}

 

poj3468，画一画然后把和式改写一下会把问题降维，使前缀和数组的前缀和的和（orz）只与单个下标有关，就用树状数组维护了。

#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;
#define ri readint()
#define gc getchar()
#define ll long long
#define maxn 100005

int readint() {
    int x = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)    x = x * 10 + c - 48;
    return x * s;
}

int N, Q;
ll sum[maxn], b[maxn][2];

ll ask(int x, int flag) {
    ll ret = 0ll;
    for (; x; x -= x & -x)    ret += b[x][flag];
    return ret;
}

void add(int x, int cnt, int flag) {
    for (; x <= N; x += x & -x)    b[x][flag] += cnt;
}

int main() {
    N = ri, Q = ri;
    for (int i = 1; i <= N; i++) {
        int a = ri;
        sum[i] = sum[i-1] + a;
    }
    for (int i = 1; i <= Q; i++) {
        char ch;
        scanf("%c", &ch);
        int a = ri, y = ri;
        if (ch == 'Q') {
            printf("%lld\n", sum[y] - sum[a-1] + (y+1)*ask(y, 0) - ask(y, 1) - a*ask(a-1, 0) + ask(a-1, 1));
        } else {
            int c = ri;
            add(a, c, 0);
            add(a, a*c, 1);
            add(y+1, -c, 0);
            add(y+1, (y+1)*(-c), 1);
        }
    }
    return 0;
}

 

hihocoder1384，读题好吃力。倍增贪心选取。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define maxn 500005
using namespace std;

int n, m, T, tmp;
ll k, a[maxn], b[maxn], c[maxn];

void merge(int l, int mid, int r) {
    for (int i = l, j = mid+1, k = l; k <= r; k++) {
        if (j > r || (i <= mid && b[i] <= b[j]))    c[k] = b[i++];
        else    c[k] = b[j++];
    }
}

ll cal(int l, int r) {
    if (r > n)    r = n;
    for (int i = tmp+1; i <= r; i++)    b[i] = a[i];
    sort(b+tmp+1, b+r+1);
    merge(l, tmp, r);

    int t = min(m, (r-l+1)/2);
    ll ret = 0ll;
    for (int i = 0; i < t; i++)
        ret += (c[r-i] - c[l+i]) * (c[r-i] - c[l+i]);
    return ret;
}

int solve() {
    int l = 1, r = 1, ans = 0;
    tmp = 1;
    b[1] = a[1];
    for (; l <= n; l = r+1, ans++) {
        int p = 1;
        while (p) {
            ll num = cal(l, r+p);

            if (num <= k) {
                tmp = r = min(r+p, n);
                for (int i = l; i <= r; i++)    b[i] = c[i];
                p <<= 1;
            } else    p >>= 1;

            if (r == n)    break;
        }
    }

    return ans;
}

int main() {
    for (scanf("%d", &T); T; T--) {
        scanf("%d%d%lld", &n, &m, &k);
        for (int i = 1; i <= n; i++)
            scanf("%lld", &a[i]);
        printf("%d\n", solve());
    }
    return 0;
}

 

UVA11235，序列过于特殊，一大堆信息都能记录，然后可以套ST表。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1e5 + 5;
const int maxnlog = 20;
struct RMQ {
    int d[maxn][maxnlog];

    void init(int *A, int n) {
        for (int i = 1; i <= n; i++)    d[i][0] = A[i];
        for (int j = 1; (1<<j) <= n; j++)
            for (int i = 1; i+(1<<j)-1 <= n; i++)
                d[i][j] = max(d[i][j-1], d[i+(1<<(j-1))][j-1]);
    }

    int query(int l, int r) {
        int k = 0;
        while ((1<<(k+1)) <= r - l + 1)    k++;
        return max(d[l][k], d[r-(1<<k)+1][k]);
    }
}rmq;

int n, q, a[maxn];
int start, t;
int cnt[maxn], num[maxn], left[maxn], right[maxn];

int main() {
    while (~scanf("%d", &n) && n) {
        scanf("%d", &q);
        for (int i = 1; i <= n; i++)    scanf("%d", &a[i]);
        a[n+1] = a[n] + 1;
        t = 0;
        for (int i = 1; i <= n+1; i++) {
            if (i == 1 || a[i] > a[i-1]) {
                if (i > 1) {
                    cnt[++t] = i - start;
                    for (int j = start; j < i; j++) {
                        num[j] = t;
                        left[j] = start;
                        right[j] = i - 1;
                    }
                }
                start = i;
            }
        }

        rmq.init(cnt, t);
        while (q--) {
            int a, b, ans;
            scanf("%d%d", &a, &b);
            if (num[a] == num[b])    ans = b - a + 1;
            else {
                ans = max(right[a] - a + 1, b - left[b] + 1);
                if (num[a]+1 < num[b])    ans = max(ans, rmq.query(num[a]+1, num[b]-1));
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}

 

poj2182，本日做的最妙的题？树状数组是维护前缀和，但前缀和得自己琢磨好。然鹅最初的最初要把题目的内在本质探索明白才行。

#include <cstdio>
#include <cmath>
#define maxn 8005

int n, a[maxn];
int c[maxn], h[maxn];

void add(int x) {
    for (; x <= n; x += x&-x)    c[x]--;
}

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        c[i]++;
        if (i + (i & -i) <= n)    c[i + (i & -i)] += c[i];
    }
    a[1] = 1;
    for (int i = 2; i <= n; i++) {
        scanf("%d", &a[i]);
        a[i]++;
    }

    int t = log(n) / log(2);
    for (int i = n; i; i--) {
        int ans = 0, sum = 0;
        for (int j = t; ~j; j--) {
            int p = 1<<j;
            if (ans + p <= n && sum + c[ans+p] < a[i]) {
                ans += p;
                sum += c[ans];
            }
        }
        add(h[i] = ans+1);
    }

    for (int i = 1; i <= n; i++)    printf("%d\n", h[i]);
    return 0;
}