Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 101173 | Accepted: 31557 | |
Case Time Limit: 2000MS |
Description
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.
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 Input
10 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
Sample Output
4 55 9 15
真的 做的我想哭
#include<stdio.h> #include<string.h> using namespace std; const int maxn = 100006; int a[maxn]; long long ans; struct node { long long data; int l,r; long long add; }tree[maxn<<2]; void build( int v, int l, int r) { tree[v].l = l; tree[v].r = r; tree[v].add = 0; if( l == r) { tree[v].data = a[r]; return; } int mid = ( l + r) / 2; build( v*2, l, mid); build( v*2+1, mid+1, r); tree[v].data = tree[2*v].data + tree[2*v+1].data; } void update( int v, int l, int r, int m) { if(tree[v].l == l && tree[v].r == r) { tree[v].data += ( r - l + 1)*m; tree[v].add += m; return; } if(tree[v].add) { tree[2*v].data+= ( tree[2*v].r - tree[2*v].l + 1)*tree[v].add; tree[2*v].add += tree[v].add; tree[2*v+1].data+= ( tree[2*v+1].r - tree[2*v+1].l + 1)*tree[v].add; tree[2*v+1].add += tree[v].add; tree[v].add = 0; } int mid = ( tree[v].l + tree[v].r ) / 2; if( r <= mid) update( v * 2,l, r, m); else { if( l > mid) update( v * 2 + 1, l, r, m); else { update( v*2, l, mid, m); update( v*2+1, mid+1, r, m); } } } void query( int v, int l, int r) { if(tree[v].l == l && tree[v].r == r) { ans += tree[v].data; return; } if(tree[v].add) { tree[2*v].data+= ( tree[2*v].r - tree[2*v].l + 1)*tree[v].add; tree[2*v].add += tree[v].add; tree[2*v+1].data+= ( tree[2*v+1].r - tree[2*v+1].l + 1)*tree[v].add; tree[2*v+1].add += tree[v].add; tree[v].add = 0; } int mid = ( tree[v].l + tree[v].r)/2; if( r <= mid ) query(v*2, l, r); else { if( l > mid ) query( v*2+1, l, r); else { query(2*v, l ,mid); query(2*v+1, mid+1, r); } } } int main() { int n,m; int i,j; while(~scanf("%d%d", &n, &m)) { for( i = 1; i <= n; i++) { scanf("%d" ,&a[i]); } build(1, 1, n); int l,r,w; char s[10]; while(m--) { scanf("%s", s); if(s[0] == 'Q') { ans = 0; scanf("%d%d", &l,&r); query(1, l, r); printf("%lld\n", ans); } else { scanf("%d%d%d", &l,&r,&w); update(1, l, r, w); } } } }