A Simple Problem with Integers(线段树 成段更新)

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 102892 Accepted: 32121Case 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.

线段树功能:update:成段增减 query:区间求和

#include <stdio.h>#define ll long long#define size 110000#define lson l, m, rt<<1#define rson m+1, r, rt<<1|1long long sum[size<<2], add[size<<2];void pushup( int rt){    sum[rt] = sum[rt<<1] + sum[rt<<1|1];}void pushdown(int rt, int m){    if(add[rt])    {        add[rt<<1] += add[rt];        add[rt<<1|1] += add[rt];        sum[rt<<1] += ( m - (m>>1)) * add[rt];        sum[rt<<1|1] += (m>>1) * add[rt];        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 updata( int L, int R, int c, int l, int r, int rt){    if( L <= l && r <= R)    {        add[rt] += c;        sum[rt] += (ll)c * ( r - l + 1);return;    }    pushdown(rt, r - l + 1);    int m = (l + r) >> 1;    if( L <= m) updata(L, R, c, lson);    if( R > m)  updata(L, R, c, rson);    pushup(rt);}long long qurey( int L, int R, int l, int r, int rt){    if( L <= l && r <= R )    {        return sum[rt];    }    pushdown(rt, r - l + 1);    int m = ( l + r) >> 1;    ll ans = 0;    if( L <= m) ans += qurey(L, R, lson);    if( R > m)  ans += qurey(L, R, rson);    return ans;}int main(){    int n, m;    char s[10];int a, b, c;    scanf("%d%d", &n, &m);        build(1, n, 1);        while(m--)        {          scanf("%s", s);          if(s[0] == 'Q')          {              scanf("%d%d",&a, &b);              printf("%lld\n",qurey(a, b, 1, n, 1));          }          else          {              scanf("%d%d%d",&a,&b,&c);updata(a, b, c, 1, n, 1);          }        }}