A Simple Problem with Integers POJ

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 117033 Accepted: 36387Case 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

代码还是源自：博客
还是附上原博主代码风格：

• lson和rson分辨表示结点的左儿子和右儿子,由于每次传参数的时候都固定是这几个变量,所以可以用预定于比较方便的表示
• 以前的写法是另外开两个个数组记录每个结点所表示的区间,其实这个区间不必保存,一边算一边传下去就行,只需要写函数的时候多两个参数,结合lson和rson的预定义可以很方便
• PushUP(int rt)是把当前结点的信息更新到父结点
• PushDown(int rt)是把当前结点的信息更新给儿子结点
• rt表示当前子树的根(root),也就是当前所在的结点

这题是两三天前做的，当时不太懂那个区间成段更新时 lazy 的思想。
其实lazy就是在更新的时候只更新到你所选取的区间，不继续向下（向子区间）更新，但是保留住当前需要继续加的值。
在下次需要更新时再更新。

#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#define inf 0x3f3f3f#define ms(x) memset(x, 0, sizeof(x))using namespace std;#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1#define ll long longconst int N = 100007;ll add[N<<2];ll sum[N<<2];void PushUp(int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void PushDown(int rt, int m){    if(add[rt])                                 //就是这里 也可能算上了之前应该更新的    {        add[rt<<1] += add[rt];        add[rt<<1|1] += add[rt];        sum[rt<<1] +=add[rt] * (m - (m>>1));        sum[rt<<1|1] += add[rt] * (m>>1);        add[rt] = 0;    }}void build(int l, int r, int rt){    add[rt] = 0;    if(l == r)    {        scanf("%lld", &sum[rt]);        return ;    }    int m = (l+r) >> 1;    build(lson);    build(rson);    PushUp(rt);}void Update(int x, int y, int c, int l, int r, int rt){    if(x<=l&&r<=y)    {        add[rt] += c;        sum[rt] +=(ll) c * (r-l+1);              //为节省时间        return ;                                 //区间更新到你所要求的地方就返回，不再向下更新。如果下次你所要求的区间更小，那么会继续更新}    PushDown(rt, r-l+1);    int m = (l + r)>>1;    if(x <= m) Update(x, y, c, lson);    if(m <  y) Update(x,y,c,rson);    PushUp(rt);}ll query(int x, int y, int l, int r, int rt){    if(x<=l && r<=y)    {        return sum[rt];    }    PushDown(rt, r-l+1);    int m = (l + r)>>1;    ll ret = 0;    if(x <= m)    ret += query(x, y, lson);    if(m < y)   ret += query(x, y, rson);    return ret;}int main(){    int n, q;    scanf("%d%d", &n, &q);    build(1, n, 1);    while(q--)    {        char s[2];        int a, b, c;        scanf("%s",s);        if(s[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;}