poj 3468 A Simple Problem with Integers

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 70937   Accepted: 21874
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

Hint

The sums may exceed the range of 32-bit integers.
题目链接：http://poj.org/problem?id=3468
线段树成段更新，成段求和。
要注意的就是可能会超过int，要用long long 
 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <string>
 5 #include <iomanip>
 6 using namespace std;
 7 #define maxn 100010
 8 #define lson l, m, rt<<1
 9 #define rson m+1, r, rt<<1|1
10 long long sum[maxn<<2], col[maxn<<2];
11 int N, Q;
12 void PushUP(int rt){
13     sum[rt] = sum[rt<<1] + sum[rt<<1|1];
14 }
15 void PushDown(int rt, int len){
16     if(col[rt]){
17         col[rt<<1] += col[rt];
18         col[rt<<1|1] += col[rt];
19         sum[rt<<1] += col[rt]*(len-(len>>1));
20         sum[rt<<1|1] += col[rt]*(len>>1);
21         col[rt] = 0;
22     }
23 }
24 void build(int l, int r, int rt){
25     col[rt] = 0;
26     if(l == r){
27         scanf("%lld", &sum[rt]);
28         return;
29     }
30     int m = (r+l)>>1;
31     build(lson);
32     build(rson);
33     PushUP(rt);
34 }
35 void update(int L, int R, int c, int l, int r, int rt){
36     if(L <= l && r <= R){
37         col[rt] += c;
38         sum[rt] += c*(r-l+1);
39         return;
40     }
41     PushDown(rt, r-l+1);
42     int m = (l+r)>>1;
43     if(L <= m) update(L, R, c, lson);
44     if(R > m) update(L, R, c, rson);
45     PushUP(rt); 
46 }
47 long long  query(int L, int R, int l, int r, int rt){
48     if(L <= l && r <= R){
49         return sum[rt];
50     }
51     PushDown(rt, r-l+1);
52     int m = (l+r)>>1;
53     long long  ret = 0;
54     if(L <= m) ret += query(L, R, lson);
55     if(m < R) ret += query(L, R, rson);
56     return ret; 
57 }
58 int main(){
59     while(~scanf("%d%d", &N, &Q)){
60         build(1, N, 1);
61         while(Q--){
62             char ch; cin>>ch;
63             int a, b, c;
64             if(ch == 'Q'){
65                 scanf("%d%d", &a, &b);
66                 printf("%lld\n",query(a,b,1,N,1));
67             }
68             else if(ch == 'C'){
69                 scanf("%d%d%d", &a, &b, &c);
70                 update(a, b, c, 1, N, 1);
71             }
72         }
73     }
74     
75     return 0;
76 }

 

posted @ 2015-04-24 17:19  下周LGD该赢了吧  阅读(107)  评论(0)  编辑  收藏  举报