POJ 3468 A Simple Problem with Integers (线段树)

题目类型  线段树 - 区间修改

题目意思

给出最多100000个数 现在有最多100000个操作

操作1  把 区间 [Ｌ,Ｒ]中的数加上一个数c

操作2  询问区间 [Ｌ,Ｒ]中的数的和是多少

解题方法

区间修改的线段树

注意懒惰标记的使用就行了

参考代码 - 有疑问的地方在下方留言 看到会尽快回复的

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define ls (rt<<1)#define rs ((rt<<1)|1)#define mid ((l+r)>>1)typedef long long LL;const int maxn = 1e5 + 10;LL sum[maxn*4], add[maxn*4];int A[maxn];void build(int rt, int l, int r) {if(l == r) {sum[rt] = A[l];return ;}build(ls, l, mid);build(rs, mid + 1, r);sum[rt] = sum[ls] + sum[rs];}void pushdown(int rt, int l, int r) {if(l != r) {if(add[rt]) {sum[ls] += (mid - l + 1) * add[rt]; add[ls] += add[rt];sum[rs] += (r - mid) * add[rt]; add[rs] += add[rt];add[rt] = 0;}}}void update(int rt, int l, int r, int L, int R, int c) {pushdown(rt, l, r);if(l == L && r == R) {sum[rt]+= (r-l+1)*c;add[rt] = c;return ;}if(R <= mid) update(ls, l, mid, L, R, c);else if(L > mid) update(rs, mid + 1, r, L, R, c);else update(ls, l, mid, L, mid, c), update(rs, mid + 1, r, mid + 1, R, c);sum[rt] = sum[ls] + sum[rs];}LL query(int rt, int l, int r, int L, int R) {pushdown(rt, l, r);if(l == L && r == R) return sum[rt];if(R <= mid) return query(ls, l, mid, L, R);else if(L > mid) return query(rs, mid + 1, r, L, R);else return query(ls, l, mid, L, mid) + query(rs, mid + 1, r, mid + 1, R);sum[rt] = sum[ls] + sum[rs];}int main() {freopen("in", "r", stdin);int n, m;while(scanf("%d%d", &n, &m) != EOF) {for( int i=0; i<n; i++ ) scanf("%d", &A[i]);build(1, 0, n);for( int i=0; i<m; i++ ) {char str[10];scanf("%s", str);if(str[0] == 'Q') {int l, r;scanf("%d%d", &l, &r);LL ans = query(1, 0, n, l-1, r-1);cout<<ans<<endl;}else {int l, r, c;scanf("%d%d%d", &l, &r, &c);update(1, 0, n, l-1, r-1, c);}}}return 0;}