PKU A Simple Problem with Integers 3468

 

代碼來自網上（非原創）

 

 

/*題目地址
對於Q操作，求出區間的總和，對於C操作，將區間裏的所有數都加上同一個數
考慮結點存放兩個信息，一個是該結點區間內的和s，一個是該區間所有數的增量d,對於一個C操作，如果區間剛好等於某結點的區間，則直接將增量加上去，否則，遞歸到左右子結點，並更新父結點的s值。這樣，對於插入與查詢都能達到O(logn)的複雜度.
Memory: 6760K
Time: 1579MS  */

#include <iostream>
using namespace std;
#define ll long long
const int MAXN = 100001;
struct {
    int l, r;
    ll s, d;
} nod[MAXN*3];
int a[MAXN];
void creat(int t, int l, int r) {
    if(l == r) {
        nod[t].l = nod[t].r = l, nod[t].s = a[l], nod[t].d = 0;
        return;
    }
    int m = (l+r) / 2;
    nod[t].l = l, nod[t].r = r, nod[t].d = 0;
    creat(t*2, l, m), creat(t*2+1, m+1, r);
    nod[t].s = nod[t*2].s + nod[t*2+1].s;
}
void inc(int t, int l, int r, int c) {
    if(l == nod[t].l && r == nod[t].r) {nod[t].d += c; return;}
    if(r <= nod[t*2].r) inc(t*2, l, r, c);
    else if(l >= nod[t*2+1].l) inc(t*2+1, l, r, c);
    else inc(t*2, l, nod[t*2].r, c), inc(t*2+1, nod[t*2+1].l, r, c);
    nod[t].s = nod[t*2].s + nod[t*2].d * (nod[t*2].r-nod[t*2].l+1) + nod[t*2+1].s + nod[t*2+1].d * (nod[t*2+1].r-nod[t*2+1].l+1);
}
ll query(int t, int l, int r) {
    if(nod[t].l == l && nod[t].r == r) return nod[t].s + nod[t].d * (r-l+1);
    ll sum;
    if(r <= nod[t*2].r) sum = query(t*2, l, r);
    else if(l >= nod[t*2+1].l) sum = query(t*2+1, l, r);
    else sum = query(t*2, l, nod[t*2].r) + query(t*2+1, nod[t*2+1].l, r);
    return sum + nod[t].d * (r-l+1);
}
int main() {
    int i, n, q, x1, x2, c;
    char s[2];
    while(scanf("%d%d", &n, &q) != EOF) {
        for(i = 1; i <= n; i++) scanf("%d", &a[i]);
        creat(1, 1, n);
        while(q--) {
            scanf("%s%d%d", s, &x1, &x2);
            if(s[0] == 'C') {scanf("%d", &c); inc(1, x1, x2, c);}
            else printf("%lld\n", query(1, x1, x2));
        }
    }
    return 0;
}