树状数组学习

树状数组是解决增加和查询的较好的方法时间复杂度均为logn

基础模板：

int lowbit(int x) {	//求x对应的2的k次方（k为x转换为2进制最低位开始连续0的个数）
	return x & (-x);
} 
void update(int i, int k) {//在i位置加k
	while(i <= N) {
		c[i] += k;
		i += lowbit(i);
	}
}
int getsum(int i) {//获得1-i数组元素的和
	int res = 0;
	while(i > 0) {
		res += c[i];
		i -= lowbit(i);
	}
	return res;
}

 来道裸题

HDU-1166

单点更新，区间查询

#include <iostream>
#include <cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
using namespace std;
const int MAXN = 50010;
int a[MAXN];
int c[MAXN];
int N;
int lowbit(int x) {	//求x对应的2的k次方（k为x转换为2进制最低位开始连续0的个数）
	return x & (-x);
} 
void update(int i, int k) {//在i位置加k
	while(i <= N) {
		c[i] += k;
		i += lowbit(i);
	}
}
int getsum(int i) {//获得1-i数组元素的和
	int res = 0;
	while(i > 0) {
		res += c[i];
		i -= lowbit(i);
	}
	return res;
}

int main () {
	int T;
	scanf("%d", &T);
	int num = 0;
	while(T--) {
		printf("Case %d:
", ++num);
		memset(a, 0, sizeof a);
		memset(c, 0, sizeof c);
		scanf("%d", &N);
		for(int i = 1; i <= N; ++i) {
			scanf("%d", &a[i]);
			update(i, a[i]);
		}
		string op;
		while(cin >> op && op != "End") {
			if(op == "Query") {
				int x, y;
				scanf("%d%d", &x, &y);
				printf("%d
", getsum(y) - getsum(x - 1));
			}else if(op == "Add") {
				int x, y;
				scanf("%d%d", &x, &y);
				update(x, y);
			}else if(op == "Sub") {
				int x, y;
				scanf("%d%d", &x, &y);
				update(x, -y);
			}
		}
	}
}

 单点查询，区间更新

解决方法：差分建树！！！！！！！！！！！！

来道裸题

落谷3368

#include <iostream>
#include <cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
using namespace std;
const int MAXN = 500010;
int a[MAXN];
int c[MAXN];
int N;
int lowbit(int x) {	//求x对应的2的k次方（k为x转换为2进制最低位开始连续0的个数）
	return x & (-x);
} 
void update(int i, int k) {//在i位置加k
	while(i <= N) {
		c[i] += k;
		i += lowbit(i);
	}
}
int getsum(int i) {//获得1-i数组元素的和
	int res = 0;
	while(i > 0) {
		res += c[i];
		i -= lowbit(i);
	}
	return res;
}

int main () {
	int M;
	scanf("%d%d", &N, &M);
	for(int i = 1; i <= N; ++i) {
		scanf("%d", &a[i]);
		update(i, a[i] - a[i - 1]);
	}
	while(M--) {
		int op;
		scanf("%d", &op);
		if(op == 1) {
			int x, y, k;
			scanf("%d%d%d", &x, &y, &k);
			update(x, k);
			update(y + 1, -k);
		}else if(op == 2) {
			int x;
			scanf("%d", &x);
			printf("%d
", getsum(x));
		}
	}

}

 来个最终版

区间更新，区间查询，如何解决？？？？？？？？

维护两个树状数组

POJ3468

#include <iostream>
#include <cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
const int MAXN = 100005;
ll a[MAXN];
ll sum1[MAXN];
ll sum2[MAXN];
int N;
ll lowbit(ll i) {
	return i & (-i);
}
void update(int i, ll k) {
	ll x = i;
	while(i <= N) {
		sum1[i] += k;
		sum2[i] += k * (x - 1ll);
		i += lowbit(i);
	}
}
ll getsum(int i) {
	ll ans = 0;
	ll x = i;
	while(i > 0) {
		ans += x * sum1[i] - sum2[i];
		i -= lowbit(i); 
	}
	return ans;
}
int main () {
	int Q;
	scanf("%d%d", &N, &Q);
	for(int i = 1; i <= N; ++i) {
		scanf("%lld", &a[i]);
		update(i, a[i] - a[i - 1]);
	}
	// cout << Q << endl;
	while(Q--) {
		char op;
		getchar();
		scanf("%c", &op);
		if(op == 'C') {
			int a, b;
			ll c;
			scanf("%d%d%lld", &a, &b, &c);
			update(a, c);
			update(b + 1, -c);
		}else if(op == 'Q') {
			int a, b;
			scanf("%d%d", &a, &b);
			printf("%lld
", getsum(b) - getsum(a - 1));
		}
	}
}