pku3468: A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 44408 Accepted: 12993Case 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

#include <iostream>#include <cstdio>using namespace std;struct node{    int l, r;    __int64 sum, d;}v[500000];__int64 A[200000];int N, Q;//往上更新void pushup(int n){    v[n].sum = v[2*n].sum + v[2*n+1].sum;}//往下更新void pushdown(int n){    if (v[n].d != 0)    {        v[n*2].d+=v[n].d;        v[n*2].sum+=(v[n*2].r-v[n*2].l+1)*v[n].d;        v[n*2+1].d+=v[n].d;        v[n*2+1].sum+=(v[n*2+1].r-v[n*2+1].l+1)*v[n].d;        v[n].d=0;    }}//建树void build(int l, int r, int n){    v[n].l = l;    v[n].r = r;    v[n].d = 0;    if (r == l)    {        v[n].sum = A[r];        return ;    }    int mid = (l+r)/2;    build(l, mid, 2*n);    build(mid+1, r, 2*n+1);    pushup(n);}void update(int l, int r, int d, int n){    if (l<=v[n].l && v[n].r<=r)    {        v[n].d += d;        v[n].sum += (v[n].r - v[n].l + 1)*d;        return ;    }    pushdown(n);    int m = (v[n].l+v[n].r)/2;    if (r <= m)        update(l, r, d, 2*n);    else if (l > m)        update(l, r, d, 2*n+1);    else    {        update(l, m, d, 2*n);        update(m+1, r, d, 2*n+1);    }    pushup(n);}__int64 query(int l, int r, int n){    if (l<=v[n].l && v[n].r<=r)        return v[n].sum;    pushdown(n);    int m = (v[n].l + v[n].r)/2;    if (r <= m)        return query(l, r, 2*n);    else if (l > m)        return query(l, r, 2*n+1);    else        return query(l, m, 2*n)+query(m+1, r, 2*n+1);}int main(){    char ch;    int a, b, c;    while (~scanf("%d %d", &N, &Q))    {        for (int i=1; i<=N; i++)            scanf("%I64d", &A[i]);        build(1, N, 1);        while (Q--)        {            getchar();            scanf("%c", &ch);            if (ch == 'Q')            {                scanf("%d%d", &a, &b);                printf("%I64d\n", query(a, b, 1));            }            else            {                scanf("%d%d%d", &a, &b, &c);                update(a, b, c, 1);            }        }    }    return 0;}