poj 3468 A Simple Problem with Integers
2013-03-04 19:14
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A Simple Problem with Integers
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
Sample Output
Hint
The sums may exceed the range of 32-bit integers.
更新区间的线段树。敲了好几遍,差不多熟悉了,其实敲多了也就明白了。
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions:40260 | Accepted: 11693 | |
Case Time Limit: 2000MS |
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.
#include <cstdio> #include <algorithm> using namespace std; #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 #define LL long long int const int maxn = 111111; LL col[maxn<<2], sum[maxn<<2]; void PushUP(int rt){ sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void PushDown(int rt, int m){ if (col[rt]){ col[rt<<1] += col[rt]; col[rt<<1|1] += col[rt]; sum[rt<<1] += (LL)(col[rt] *(m-(m>>1))); sum[rt<<1|1] += (LL)(col[rt] *(m>>1)); col[rt] = 0; } } void build(int l, int r, int rt){ col[rt] = 0; if (l == r) {scanf("%lld", sum + rt); return;} int m = (l + r) >> 1; build(lson); build(rson); PushUP(rt); } void update(int L, int R, int c, int l, int r, int rt){ if (L <= l && R >= r) {col[rt] += c; sum[rt] += (LL)(c*(r-l+1)); return;} PushDown(rt, r - l + 1); int m= (l + r) >> 1; if (L <= m) update(L, R, c, lson); if (R > m) update(L, R, c, rson); PushUP(rt); } LL query(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 ret = 0; if (L <= m) ret += query(L, R, lson); if (R > m) ret += query(L, R, rson); PushUP(rt); return ret; } int main(void){ int n, q; #ifndef ONLINE_JUDGE freopen("poj3468.in", "r", stdin); #endif while (~scanf("%d%d", &n, &q)){ build(1, n, 1); while (q--){ char a[2]; scanf("%s", a); int m, b, c; if (a[0] == 'Q') {scanf("%d%d", &m, &b);printf("%lld\n",query(m, b, 1, n, 1));} else { scanf("%d%d%d", &m, &b, &c); update(m, b, c, 1, n, 1); } } } return 0; }
更新区间的线段树。敲了好几遍,差不多熟悉了,其实敲多了也就明白了。
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