poj3468 A Simple Problem with Integers 线段树区间更新

A Simple Problem with Integers
Time Limit: 5000MS   Memory Limit: 131072K
Total Submissions: 97722   Accepted: 30543
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.
  1 #include <iostream>
  2 #include <sstream>
  3 #include <fstream>
  4 #include <string>
  5 #include <vector>
  6 #include <deque>
  7 #include <queue>
  8 #include <stack>
  9 #include <set>
 10 #include <map>
 11 #include <algorithm>
 12 #include <functional>
 13 #include <utility>
 14 #include <bitset>
 15 #include <cmath>
 16 #include <cstdlib>
 17 #include <ctime>
 18 #include <cstdio>
 19 #include <cstring>
 20 #define FOR(i, a, b)  for(int i = (a); i <= (b); i++)
 21 #define RE(i, n) FOR(i, 1, n)
 22 #define FORP(i, a, b) for(int i = (a); i >= (b); i--)
 23 #define REP(i, n) for(int i = 0; i <(n); ++i)
 24 #define SZ(x) ((int)(x).size )
 25 #define ALL(x) (x).begin(), (x.end())
 26 #define MSET(a, x) memset(a, x, sizeof(a))
 27 using namespace std;
 28 
 29 
 30 typedef long long int ll;
 31 typedef pair<int, int> P;
 32 int read() {
 33     int x=0,f=1;
 34     char ch=getchar();
 35     while(ch<'0'||ch>'9') {
 36         if(ch=='-')f=-1;
 37         ch=getchar();
 38     }
 39     while(ch>='0'&&ch<='9') {
 40         x=x*10+ch-'0';
 41         ch=getchar();
 42     }
 43     return x*f;
 44 }
 45 const double pi=3.14159265358979323846264338327950288L;
 46 const double eps=1e-6;
 47 const int mod = 1e9 + 7;
 48 const int INF = 0x3f3f3f3f;
 49 const int MAXN = 100003;
 50 const int xi[] = {0, 0, 1, -1};
 51 const int yi[] = {1, -1, 0, 0};
 52 
 53 int N, T;
 54 struct seg {
 55     ll sum, add;
 56 } a[MAXN<<2];
 57 ll b[MAXN];
 58 
 59 void pushdown(int k, int m){
 60     if(a[k].add){
 61         a[k<<1].add += a[k].add;
 62         a[k<<1|1].add += a[k].add;
 63         a[k<<1].sum += a[k].add*(m - (m>>1));
 64         a[k<<1|1].sum += a[k].add*(m>>1);
 65         a[k].add = 0;
 66     }
 67 }
 68 void build(int k, int l, int r) {
 69     a[k].add = 0;
 70     if(r == l) {
 71         a[k].sum = b[l];
 72         return;
 73     }
 74     int m = (l+r)>>1;
 75     build(k<<1, l, m);
 76     build(k<<1|1, m+1, r);
 77     a[k].sum = a[k<<1].sum + a[k<<1|1].sum;
 78 }
 79 void update(int k, int x, int add, int l, int r) {
 80     if(l == r ) {
 81         a[k].sum += add;
 82         return;
 83     }
 84     int m = (l + r)>>1;
 85     if(x <= m) {
 86         update(k<<1, x, add, l, m);
 87     } else update(k<<1|1, x, add, m+1, r);
 88     a[k].sum = a[k<<1].sum + a[k<<1|1].sum;
 89 }
 90 void change(int k, int l, int r, int ca, int cb, ll c) {
 91     if(ca <= l && cb >= r){
 92         a[k].add += c;
 93         a[k].sum += c*(r - l + 1);
 94         return;
 95     }
 96     pushdown(k, r-l+1);
 97     int m = (l+r) >>1;
 98     if(ca <= m) change(k<<1, l, m, ca, cb, c);
 99     if(cb > m) change(k <<1|1, m+1, r, ca, cb, c);
100     a[k].sum = a[k<<1].sum + a[k<<1|1].sum;
101 }
102 ll query(int k, int l, int r, int qa, int qb) {
103     if(qa <= l && qb >= r) return a[k].sum;
104     pushdown(k, r-l+1);
105     int m = (l+r) >>1;
106     ll ans = 0;
107     if(qa <= m) ans += query(k<<1, l, m, qa, qb);
108     if(qb > m) ans += query(k<<1|1, m+1, r, qa, qb);
109     return ans;
110 }
111 
112 int main() {
113     //freopen("in.txt", "r", stdin);
114     int n, Q;
115     scanf("%d%d", &n, &Q);
116     for(int i = 1; i <= n; i++) {
117         scanf("%I64d", &b[i]);
118     }
119     build(1, 1, n);
120     char s[10];
121     int ca, cb;
122     while(Q--) {
123         scanf("%s%d%d", s, &ca, &cb);
124         if(s[0] == 'Q') {
125             printf("%I64d\n", query(1, 1, n, ca, cb));
126         }
127         else {
128             ll add;
129             scanf("%I64d", &add);
130 //            printf("SSS\n");
131             change(1, 1, n, ca, cb, add);
132 //            printf("SSS\n");
133         }
134     }
135 
136     return 0;
137 }

 

 

posted on 2016-09-30 21:46  disppr  阅读(171)  评论(0)  编辑  收藏  举报