poj 3468 A Simple Problem with Integers （线段树 成段更新 加值 求和）

题目链接

题意：

只有这两种操作

C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

代码风格更新后：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <algorithm>
 #define lson l, mid, 2*rt
 #define rson mid+1, r, 2*rt+1
 #define LL __int64
 const int maxn = +;
 using namespace std;
 LL sum[*maxn], lz[*maxn];

 void pushup(int rt)
 {
     sum[rt] = sum[*rt]+sum[*rt+];
 }
 void pushdown(int l, int r, int rt)
 {
     if(lz[rt]!=)
     {
         int mid = (l+r)/;
         sum[*rt] += lz[rt]*(mid-l+);
         sum[*rt+] += lz[rt]*(r-mid);
         lz[*rt] += lz[rt];
         lz[*rt+] += lz[rt];
         lz[rt] = ;
     }
 }
 void build(int l, int r, int rt)
 {
     if(l==r)
     {
         scanf("%I64d", &sum[rt]);
         return;
     }
     int mid = (l+r)/;
     build(lson);
     build(rson);
     pushup(rt);
 }
 void update(int ll, int rr, LL c, int l, int r, int rt)
 {
     if(ll>r) return;
     if(rr<l) return;
     if(ll<=l && rr>=r)
     {
         lz[rt] += c;
         sum[rt] += (r-l+)*c;
         return;
     }
     pushdown(l, r, rt);
     int mid = (l+r)/;
     update(ll, rr, c, lson);
     update(ll, rr, c, rson);
     pushup(rt);
 }
 LL query(int ll, int rr, int l, int r, int rt)
 {
     if(ll>r) return ;
     if(rr<l) return ;
     if(ll<=l && rr>=r)
         return sum[rt];
     pushdown(l, r, rt);
     int mid = (l+r)/;
    return query(ll, rr, lson)+query(ll, rr, rson);
 }
 int main()
 {
     int n, q, a, b;
     LL c;
     char ch;
     while(~scanf("%d%d", &n, &q))
     {
         memset(sum, , sizeof(sum));
         memset(lz, , sizeof(lz));
         build(, n, );
         while(q--)
         {
             getchar();
             scanf("%c %d %d", &ch, &a, &b);
             if(ch=='Q')
             printf("%I64d\n", query(a, b, , n, ));
             else
             {
                 scanf("%I64d", &c);
                 update(a, b, c, , n, );
             }
         }
     }
     return ;
 }

代码风格更新前：

分析：自己写的有点麻烦了，写的时候手残+脑残，改了好久。

val+lazy*(r-l+1)表示和，如果lazy==0表示当前区间加的值不统一。

 #include <iostream>
 #include <cstdio>
 #include <vector>
 #include <cstring>
 #include <cstdlib>
 #include <algorithm>
 #define LL __int64
 const int maxn = +;
 using namespace std;
 int n, q;
 __int64 a[maxn];
 struct line
 {
     int l, r;
     LL val, lazy;
 }tr[*maxn];

 void build(int o, int l, int r)
 {
     tr[o].l = l; tr[o].r = r;
     tr[o].lazy = ;
     if(l==r)
     {
         tr[o].val = a[l];
         return;
     }
     int mid = (l+r)/;
     build(*o, l, mid);
     build(*o+, mid+, r);
     tr[o].val = tr[*o].val+tr[*o+].val;
 }
 void update(int o, int l, int r, int add)
 {
     if(tr[o].l==l && tr[o].r==r)
     {
         tr[o].lazy += add;
         return;
     }
     if(tr[o].lazy)  //之前没向下更新的向下更新
     {
         tr[*o].lazy += tr[o].lazy;
         tr[*o+].lazy += tr[o].lazy;
         tr[o].val += (LL)(tr[o].lazy*(tr[o].r-tr[o].l+));  //同时把lazy存的加到和里
         tr[o].lazy = ;
     }
     tr[o].val += (LL)(add*(r-l+)); //在这个过程中把新增加的加上，注意左右区间
     int mid = (tr[o].l+tr[o].r)/;
     if(r <= mid) update(*o, l, r, add);
     else if(l > mid) update(*o+, l, r, add);
     else
     {
         update(*o, l, mid, add);
         update(*o+, mid+, r, add);
     }
 }

 LL query(int o, int l, int r)
 {
     if(tr[o].l==l && tr[o].r==r)
     return tr[o].val + tr[o].lazy*(r-l+);
     if(tr[o].lazy)  //由于之前可能存在 没有更新的所以向下更新
     {
         tr[*o].lazy += tr[o].lazy;
         tr[*o+].lazy += tr[o].lazy;
         tr[o].val += (LL)(tr[o].lazy*(tr[o].r-tr[o].l+));
         tr[o].lazy = ;
     }
     int mid = (tr[o].l+tr[o].r)/;
     if(r<=mid) return query(*o, l, r);
     else if(l > mid) return query(*o+, l, r);
     else
     {
        return query(*o, l, mid) + query(*o+, mid+, r);
     }
 }
 int main()
 {
     int i, l, r, add;
      char ch;
     while(~scanf("%d%d", &n, &q))
     {
         for(i = ; i <= n; i++)
         scanf("%I64d", &a[i]);
         build(, , n);
         for(i = ; i <= q; i++)
         {
             getchar();
             scanf("%c", &ch);
             if(ch=='Q')
             {
                 scanf("%d%d", &l, &r);
                 printf("%I64d\n", query(, l, r));
             }
             else
             {
                 scanf("%d%d%d", &l, &r, &add);
                 update(, l, r, add);
             }
         }
     }
     return ;
 }