POJ 3468 A Simple Problem with Integers(线段树区间修改)
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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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4
Sample Output
455915
Hint
The sums may exceed the range of 32-bit integers.
大意:对于每个C询问,将区间[a,b]上的元素都加c;对于每个Q询问,输出区间[a,b]的和。
思路:简单的线段树区间更新。
#include<cstdio>#include<cstring>#include<iostream>#include<cmath>#include<vector>#include<algorithm>#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lc rt<<1#define rc rt<<1|1using namespace std;typedef long long ll;const int maxn = 200000 + 10;const int INF = 1e9 + 10;ll sum[maxn],add[maxn],ans;void PushUp(int rt){ sum[rt] = sum[lc] + sum[rc];}void PushDown(int rt, int m){ if (add[rt]) { add[lc] += add[rt]; add[rc] += add[rt]; sum[lc] += add[rt] * (m - (m >> 1)); sum[rc] += add[rt] * (m >> 1); add[rt] = 0; }}void build(int l, int r, int rt){ if (l == r) scanf("%lld",&sum[rt]); else { int m = (l + r) >> 1; build(lson); build(rson); PushUp(rt); }}void update(int ql, int qr, int c, int l, int r, int rt){ if (ql <= l && r <= qr) { //如果完全覆盖,那么暂时不更新子结点,因为暂时不会用到,以节省时间。 add[rt] += c; sum[rt] += (ll)c*(r-l+1); return; } //如果没有完全覆盖,则要考虑左右子结点,这时要先把左右子结点更新一下再操作 PushDown(rt, r-l+1); int m = (l + r) >> 1; if (ql <= m) update(ql, qr, c, lson); if (m < qr) update(ql, qr, c, rson); //因为子结点已经被修改,最后更新父节点 PushUp(rt);}ll query(int ql, int qr, int l, int r, int rt){ if (ql <= l && r <= qr) { return sum[rt]; } PushDown(rt, r-l+1); int m = (l + r) >> 1; ll ans = 0; if (ql <= m) ans += query(ql, qr, lson); if (m < qr) ans += query(ql, qr, rson); //这里不用再更新sum[rt],因为这时的sum[rt]就是真实的区间和 return ans;}int main(){ int n,q; scanf("%d%d",&n,&q); build(1,n,1); while(q--) { char s[2]; scanf("%s",s); int a,b,c; if (s[0] == 'Q') { scanf("%d%d",&a,&b); ll ans = query(a,b,1,n,1); printf("%lld\n",ans); } else { scanf("%d%d%d",&a,&b,&c); update(a,b,c,1,n,1); } }}
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