A Simple Problem with Integers 线段树区间查询+区间修改

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 100883 Accepted: 31450Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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.

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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

Output

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

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

Hint

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly--2007.11.25, Yang Yi

简单的套用线段树的模板，用区间修改和区间查询，另外一点要注意的是，数据输出时候要用%lld, 还有就是add数组要开成 long long 的

AC代码：

#include<cstdio>#include<cstring>   #include<algorithm>#include<iostream>#include<string>#include<cmath>#include<iomanip>using namespace std;typedef long long ll;const int maxn = 100002;int n;ll sum[maxn << 2], add[maxn << 2], people[maxn];void pushup(int rt) {sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];}void build(int l, int r, int rt) {if(l == r) {sum[rt] = people[l];return;}int m = (l + r) >> 1;build(l, m, rt << 1);build(m + 1, r, rt << 1 | 1);pushup(rt);}void pushdown(int rt, int lnum, int rnum) {if(add[rt]) {add[rt << 1] += add[rt];   //这个地方一定要用+=， 因为这是一个累加的过程add[rt << 1 | 1] += add[rt];sum[rt << 1]  += add[rt] * lnum;  //这个地方乘的是add[rt]，因为这次修改的是因为他的父节点而改变的带点sum[rt << 1 | 1] += add[rt] * rnum;add[rt] = 0;}}void update(int L, int R, int C, int l, int r, int rt) {if(l >= L && r <= R) {sum[rt] += C * (r - l + 1);add[rt] += C; //同一个点可能被多次标记，所以用 +=,而不用= return;}int m = (l + r) >> 1;pushdown(rt, m - l +1, r - m);if(L <= m)  update(L, R, C, l, m, rt << 1);if(R > m)  update(L, R, C, m+1, r, rt << 1 | 1);pushup(rt);}ll query(int L, int R, int l, int r, int rt) {if(l >= L && r <= R) {return sum[rt];}int m = (r + l) >> 1;ll ans = 0;pushdown(rt, m - l + 1, r - m);if(L <= m)  ans += query(L, R, l, m, rt << 1);if(R > m)  ans += query(L, R, m + 1, r, rt << 1 | 1);return ans;}int main(){int q = 0;char que[2];scanf("%d %d",&n, &q);int i;for(i=1; i<=n; ++i)scanf("%lld",&people[i]);build(1, n, 1);int a, b, c;while(q--) {scanf("%s", que);if(que[0] == 'Q') {scanf("%d %d", &a, &b);printf("%lld\n",query(a, b, 1, n, 1));}else {scanf("%d %d %d", &a, &b, &c);update(a, b, c, 1, n, 1);}}return 0;}