A Simple Problem with Integers BZOJ3212 线段树

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Sample Output4
55
9
15

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

 

就当练练手了。。

水题一道；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/

ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}

int n;
ll a[maxn];

struct node {
	ll l, r;
	ll lazy;
	ll sum;
}tree[maxn<<1];

void pushup(int rt) {
	tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
}

void build(int l, int r, int rt) {
	tree[rt].l = l; tree[rt].r = r; tree[rt].lazy = 0;
	if (l == r) {
		tree[rt].lazy = 0;
		tree[rt].sum = a[l]; return;
	}
	int mid = (l + r) >> 1;
	build(l, mid, rt << 1); build(mid + 1, r, rt << 1 | 1);
	pushup(rt);
}

void pushdown(int rt) {
	if (tree[rt].lazy) {
		tree[rt << 1].sum += (ll)tree[rt].lazy*(tree[rt << 1].r - tree[rt << 1].l + 1);
		tree[rt << 1 | 1].sum += (ll)tree[rt].lazy*(tree[rt << 1 | 1].r - tree[rt << 1 | 1].l + 1);
		tree[rt << 1].lazy += tree[rt].lazy;
		tree[rt << 1 | 1].lazy += tree[rt].lazy;
		tree[rt].lazy = 0;
	}
}

void upd(int l, int r, int rt,ll val) {
	if (l <= tree[rt].l&&tree[rt].r <= r) {
		tree[rt].sum += (tree[rt].r - tree[rt].l + 1)*val;
		tree[rt].lazy += val; return;
	}
	pushdown(rt);
	int mid = (tree[rt].l + tree[rt].r) >> 1;
	if (l <= mid)upd(l, r, rt << 1, val);
	if (mid < r)upd(l, r, rt << 1 | 1, val);
	pushup(rt);
}

ll query(int l, int r, int rt) {
	if (l <= tree[rt].l&&tree[rt].r <= r) {
		return tree[rt].sum;
	}
	pushdown(rt);
	int mid = (tree[rt].r + tree[rt].l) >> 1;
	ll ans = 0;
	if (l <= mid)ans += query(l, r, rt << 1);
	if (mid < r)ans += query(l, r, rt << 1 | 1);
	return ans;
}

int main()
{
	//ios::sync_with_stdio(0);
	rdint(n); int q; rdint(q);
	for (int i = 1; i <= n; i++)rdllt(a[i]);
	build(1, n, 1);
	while (q--) {
		char op; int a, b;
		cin >> op;
		if (op == 'C') {
			ll v; rdint(a); rdint(b); rdllt(v);
			upd(a, b, 1, v);
		}
		else {
			rdint(a); rdint(b);
			cout << query(a, b, 1) << endl;
		}
	}
	return 0;
}