线段树

动态区间最小值

#include <iostream>
#include <cstdio>
using namespace std;

#define MAX_LEN 100010
//维护最小值的线段树 
int A[MAX_LEN];
int S[MAX_LEN]; //S[i]存储结点i的询问信息 

void Build(int o, int l, int r) { //建结点o 
	if(l == r) {
		S[o] = A[l];
		return;
	}
	int mid = l + (r - l) / 2;
	Build(o*2, l, mid);
	Build(o*2+1, mid+1, r);	
	S[o] = min(S[o*2], S[o*2+1]);
}

void Update(int o, int l, int r, int ind, int ans) { //o为当前更新的结点，ind为修改的结点 
	if(l == r) {
		S[o] = ans;
		return;
	}
	int mid = l + (r - l) / 2;
	if(ind <= mid) Update(o*2, l, mid, ind, ans);
	else Update(o*2+1, mid+1, r, ind, ans);
	S[o] = min(S[o*2], S[o*2+1]);
}

int Query(int o, int l, int r, int ql, int qr) {
	int ans = 1e9, mid = l + (r - l) / 2;
	if(ql > r || qr < l) return ans;
	if(ql <= l && r <= qr) return S[o];
	if(ql <= mid) ans = min(ans, Query(o*2, l, mid, ql, qr));
	if(qr > mid) ans = min(ans, Query(o*2+1, mid+1, r, ql, qr));
	return ans;
}

int n, m;

int main() {
	int flag, ind, k;
	scanf("%d %d", &n, &m);
	for(int i=1; i<=n; i++) scanf("%d", &A[i]);
	Build(1, 1, n); 
	for(int i=1; i<=m; i++) {
		scanf("%d%d%d", &flag, &ind, &k);
		if(flag == 1) Update(1, 1, n, ind, k);
		else if(flag == 2) printf("%d\n", Query(1, 1, n, ind, k));
	}
	return 0;
}

点修改，求区间和

#include <iostream>
#include <cstdio>
using namespace std;

#define MAX_LEN 500001
//点修改 求区间和 的线段树
int A[MAX_LEN];
int S[MAX_LEN<<2]; //S[i]存储结点i的询问信息

void Build(int o, int l, int r) { //建结点o
	if(l == r) {
		S[o] = A[l];
		return;
	}
	int mid = l + (r - l) / 2;
	Build(o*2, l, mid);
	Build(o*2+1, mid+1, r);
	S[o] = S[o*2] + S[o*2+1];
}

void Update(int o, int l, int r, int ind, int ans) { //o为当前更新的结点，ind为修改的结点
	if(l == r) {
		S[o] += ans;
		return;
	}
	int mid = l + (r - l) / 2;
	if(ind <= mid) Update(o*2, l, mid, ind, ans);
	else Update(o*2+1, mid+1, r, ind, ans);
	S[o] += ans;
}

int Query(int o, int l, int r, int ql, int qr) {
	int ans = 0, mid = l + (r - l) / 2;
	if(ql > r || qr < l) return 0;
	if(ql <= l && r <= qr) return S[o]; 
	if(ql <= mid) ans += Query(o*2, l, mid, ql, qr);
	if(qr > mid) ans += Query(o*2+1, mid+1, r, ql, qr);
	return ans;
}

int n, m;

int main() {
	int flag, ind, k;
	scanf("%d %d", &n, &m);
	for(int i=1; i<=n; i++) scanf("%d", &A[i]);
	Build(1, 1, n);
	for(int i=1; i<=m; i++) {
		scanf("%d%d%d", &flag, &ind, &k);
		if(flag == 1) Update(1, 1, n, ind, k);
		else if(flag == 2) printf("%d\n", Query(1, 1, n, ind, k));
	}
	return 0;
}

区间修改区间求和（Luogu【模版】 线段树1）

#include <iostream>
using namespace std;

typedef long long LL;

LL sum[400010], n, q;
LL add[400010], A[100010];

void build_tree(LL k1, LL l, LL r) {
    if(l == r) {
        sum[k1] = A[l];
        return;
    }
    LL mid = l+r >> 1;
    build_tree(k1<<1, l, mid);
    build_tree(k1<<1|1, mid+1, r);
    sum[k1] = sum[k1<<1] + sum[k1<<1|1];
}

void pushdown(int k1, int l, int r) {
    LL mid = l+r >> 1;
    add[k1<<1] += add[k1];
    add[k1<<1|1] += add[k1];
    sum[k1<<1|1] += add[k1] * (r-mid);
    sum[k1<<1] += add[k1] * (mid-l+1);
    add[k1] = 0;
}

void Update(LL k1, LL l, LL r, LL k, LL L, LL R) {
    if(l > R || r < L) return;
    if(L <= l && r <= R) {
        sum[k1] += k*(r-l+1);
        add[k1] += k;
        return;
    }
    LL mid = l+r >> 1;
    pushdown(k1, l, r);
    Update(k1<<1, l, mid, k, L, R);
    Update(k1<<1|1, mid+1, r, k, L, R);
    sum[k1] = sum[k1<<1] + sum[k1<<1|1];
}

LL Query(LL k1, LL l, LL r, LL L, LL R) {
    if(l > R || r < L) return 0;
    if(L <= l && r <= R) {
        return sum[k1];
    }
    LL mid = l + r >> 1;
    pushdown(k1, l, r);
    return Query(k1<<1, l, mid, L, R) + Query(k1<<1|1, mid+1, r, L, R);
}

int main() {
    cin >> n >> q;
    for(LL i=1; i<=n; i++) cin >> A[i];
    build_tree(1, 1, n);
    for(LL i=1; i<=q; i++) {
        LL opt, x, y, k;
        cin >> opt;
        if(opt == 1) {
            cin >> x >> y >> k;
            Update(1, 1, n, k, x, y);
        } else if(opt == 2) {
            cin >> x >> y;
            cout << Query(1, 1, n, x, y) << endl;
        }
    }
    return 0;
}

区间两种修改区间求和（Luogu【模版】线段树2）

#include <iostream>
#include <cstdio>
using namespace std;

typedef long long LL;
#define MAXN 100010

LL N, M, P;
LL sum[MAXN<<2], A[MAXN];
LL add[MAXN<<2], mul[MAXN<<2];

void build(LL k, LL l, LL r) {
    if(l == r) {
        sum[k] = A[l] % P;
        return;
    }
    LL mid = l+r >> 1;
    build(k<<1, l, mid);
    build(k<<1|1, mid+1, r);
    sum[k] = (sum[k<<1] + sum[k<<1|1]) % P;
    add[k] = 0;
    mul[k] = 1;
}

void Update_add(LL k1, LL l, LL r, LL x) {
    add[k1] += x;
    add[k1] %= P;
    sum[k1] = (sum[k1] + (r-l+1) * x) % P;
}

void Update_mul(LL k1, LL l, LL r, LL x) {
    add[k1] *= x;
    add[k1] %= P;
    mul[k1] *= x;
    mul[k1] %= P;
    sum[k1] = (sum[k1] * x) % P;
}

void pushdown(LL k1, LL l, LL r) {
    LL mid = l+r >> 1; 
    if(mul[k1] != 1) {
        Update_mul(k1<<1, l, mid, mul[k1]);
        Update_mul(k1<<1|1, mid+1, r, mul[k1]);
        mul[k1] = 1;
    }
    if(add[k1]) {
        Update_add(k1<<1, l, mid, add[k1]);
        Update_add(k1<<1|1, mid+1, r, add[k1]);
        add[k1] = 0;
    }
}


void ADD(LL k1, LL l, LL r, LL x, LL L, LL R) {
    if(l > R || r < L) return;
    if(L <= l && r <= R) {
        Update_add(k1, l, r, x);
        return;
    }
    LL mid = l+r >> 1;
    pushdown(k1, l, r);
    ADD(k1<<1, l, mid, x, L, R);
    ADD(k1<<1|1, mid+1, r, x, L, R);
    sum[k1] = (sum[k1<<1] + sum[k1<<1|1]) % P;
}

void MUL(LL k1, LL l, LL r, LL x, LL L, LL R) {
    if(l > R || r < L) return;
    if(L <= l && r <= R) {
        Update_mul(k1, l, r, x);
        return;
    }
    LL mid = l+r >> 1;
    pushdown(k1, l, r);
    MUL(k1<<1, l, mid, x, L, R);
    MUL(k1<<1|1, mid+1, r, x, L, R);
    sum[k1] = (sum[k1<<1] + sum[k1<<1|1]) % P;
}

LL Query(LL k1, LL l, LL r, LL L, LL R) {
    if(l > R || r < L) return 0;
    if(L <= l && r <= R) return sum[k1];
    LL mid = l+r >> 1;
    pushdown(k1, l, r);
    return (Query(k1<<1, l, mid, L, R) + Query(k1<<1|1, mid+1, r, L, R)) % P;
}

int main() {
    LL opt, x, y, k;
    scanf("%lld%lld%lld", &N, &M, &P);
    for(LL i=1; i<=N; i++) scanf("%lld", &A[i]);
    build(1, 1, N);
    for(LL i=1; i<=M; i++) {
        scanf("%lld%lld%lld", &opt, &x, &y);
        if(opt == 1) {
            scanf("%lld", &k);
            MUL(1, 1, N, k, x, y);
        } else if(opt == 2) {
            scanf("%lld", &k);
            ADD(1, 1, N, k, x, y);
        } else if(opt == 3) {
            printf("%lld\n", Query(1, 1, N, x, y));
        }
    }
    return 0;
}