A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 37225 | Accepted: 10745 | |
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.
Source
POJ Monthly--2007.11.25, Yang Yi
分析:线段树:成段增加,成段求和。
#include<cstdio> #include<cstring> #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long u64; u64 sum[1 << 18]; int col[1 << 18]; void pushup(int rt) { sum[rt] = sum[rt << 1] + sum[rt << 1 | 1]; } void pushdown(int rt, int len) { if (col[rt]) { col[rt << 1] += col[rt]; col[rt << 1 | 1] += col[rt]; sum[rt << 1] += (u64) (len - (len >> 1)) * col[rt]; sum[rt << 1 | 1] += (u64) (len >> 1) * col[rt]; col[rt] = 0; } } void build(int l, int r, int rt) { 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] += (u64) 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); } u64 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; u64 ans = 0; if (L <= m) ans += query(L, R, lson); if (R > m) ans += query(L, R, rson); return ans; } int main() { int n, m, a, b, c; char op[4]; while (scanf("%d%d", &n, &m) != EOF) { memset(col, 0, sizeof (col)); build(1, n, 1); while (m--) { scanf("%s", op); if (op[0] == 'Q') { scanf("%d%d", &a, &b); printf("%lld\n", query(a, b, 1, n, 1)); } else if (op[0] == 'C') { scanf("%d%d%d", &a, &b, &c); update(a, b, c, 1, n, 1); } } } return 0; }