题目大意：

给定一个长度为N的数列，输入Q行操作指令。指令有两种：

1.“C l r d”，区间A[l～r]都加上d

2.“Q l r”，查询区间A[l～r]每个数的和并输出

解题思路：

线段树模板啊～～～～

Accepted code：

#include
    
     #include
     
      #define ls (now<<1)#define rs (now<<1|1)#define l(x) tree[x].l#define r(x) tree[x].r#define S(x) tree[x].sum#define A(x) tree[x].lazy#define N 100010#define mid(x,y) ((x+y)>>1)using namespace std;struct Segment_Tree {    int l,r;    long long sum,lazy;}tree[N<<2];int a[N],n,m;void build(int now,int l,int r) {    l(now)=l; r(now)=r;    if (l==r) {
   scanf("%lld",&S(now));return;}    build(ls,l,mid(l,r));    build(rs,mid(l,r)+1,r);    S(now)=S(ls)+S(rs);}void down_update(int now) {    if (A(now)) {        S(ls)+=A(now)*(r(ls)-l(ls)+1); A(ls)+=A(now);        S(rs)+=A(now)*(r(rs)-l(rs)+1); A(rs)+=A(now);        A(now)=0;    }}void change(int now,int l,int r,int d) {    if (l<=l(now)&&r>=r(now)) {        S(now)+=(long long)d*(r(now)-l(now)+1);        A(now)+=d; return;    }    down_update(now); int mi=mid(l(now),r(now));    if(l<=mi) change(ls,l,r,d);    if(r>mi) change(rs,l,r,d);    S(now)=S(ls)+S(rs);}long long ask(int now,int l,int r) {    if (l<=l(now)&&r>=r(now)) return S(now);    down_update(now);    int mi=mid(l(now),r(now));    long long All=0;    if (l<=mi) All+=ask(ls,l,r);    if (r>mi) All+=ask(rs,l,r);    return All;}int main() {    scanf("%d%d",&n,&m);    build(1,1,n);    while (m--) {        char c[2];int x,y,dat;        scanf("%s%d%d",c,&x,&y);        if (c[0]=='C') {            scanf("%d",&dat);            change(1,x,y,dat);        } else printf("%lld\n",ask(1,x,y));    }}