PKU 3468 A Simple Problem with Integers

题目：http://poj.org/problem?id=3468

题意：有N个整数，有两个操作，一个是对给定的区间增加值，一个是查询给定区间的和

#include <stdio.h>
const int maxn=100000;
struct SegmentTree
{
int l, r;
long long sum, lazy;
}st[maxn<<2];
long long z[maxn+10], ans, c;
int n, q, a, b;
char ch[3];
void BuildTree(int v, int left, int right)
{
st[v].l=left, st[v].r=right, st[v].lazy=0;
if (left==right)
{
st[v].sum=z[left];
return;
}
int lc=v<<1;
int rc=lc+1;
int mid=(left+right)>>1;
BuildTree(lc, left, mid);
BuildTree(rc, mid+1, right);
st[v].sum=st[lc].sum+st[rc].sum;
}
void Insert(int left, int right, int v, long long x)
{
if (st[v].l==left&&st[v].r==right)
{
st[v].lazy+=x;
return;
}
int lc=v<<1;
int rc=lc+1;
int mid=(st[v].l+st[v].r)>>1;
st[v].sum+=(x*(right-left+1));
if (left>mid) Insert(left, right, rc, x);
else if (right<=mid) Insert(left, right, lc, x);
else
{
Insert(left, mid, lc, x);
Insert(mid+1, right, rc, x);
}
}
long long Query(int left, int right, int v)
{
if (st[v].l==left&&st[v].r==right)
{
ans+=st[v].sum+st[v].lazy*(right-left+1);
return ans;
}
int lc=v<<1;
int rc=lc+1;
int mid=(st[v].l+st[v].r)>>1;
st[v].sum+=st[v].lazy*(st[v].r-st[v].l+1);
st[lc].lazy+=st[v].lazy, st[rc].lazy+=st[v].lazy;
st[v].lazy=0;
if (left>mid) Query(left, right, rc);
else if (right<=mid) Query(left, right, lc);
else Query(left, mid, lc)+Query(mid+1, right, rc);
}
int main()
{
//freopen("in.txt", "r", stdin);
while (scanf("%d %d", &n, &q)==2)
{
for (int i=1; i<=n; i++) scanf("%lld", &z[i]);
BuildTree(1, 1, n);
for (int i=0; i<q; i++)
{
scanf("%s", ch);
if (ch[0]=='C')
{
scanf("%d %d %lld", &a, &b, &c);
Insert(a, b, 1, c);
}
else
{
scanf("%d %d", &a, &b);
ans=0;
printf("%lld\n", Query(a, b, 1));
}
}
}
return 0;
}