poj3468 A Simple Problem with Integers 【线段树】

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题目链接：A Simple Problem with Integers
题意：n个数，q次操作，操作分为询问和修改操作，修改是让[a,b]的元素加上c，查询是求[a,b]元素的和
解析：线段树，区间修改，区间查询

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int maxn = 100000+100;
struct node
{int l,r;long long lazy,sum;void update(int val){sum += 1LL*(r-l+1)*val;lazy+= val;}
}tree[4*maxn];
int a[maxn];
void push_up(int i)
{tree[i].sum = tree[i<<1].sum + tree[i<<1|1].sum;
}
void push_down(int i)
{long long lazyVal = tree[i].lazy;if(lazyVal){tree[i<<1].update(lazyVal);tree[i<<1|1].update(lazyVal);tree[i].lazy = 0;}
}
void build(int i,int l,int r)
{tree[i].l = l,tree[i].r = r;tree[i].sum = tree[i].lazy = 0;if(l==r){tree[i].sum = a[r];return ;}int mid = (l+r)>>1;build(i<<1,l,mid);build(i<<1|1,mid+1,r);push_up(i);
}
void update(int i,int l,int r,int val)
{int L = tree[i].l;int R = tree[i].r;if(l<=L && R<=r){tree[i].update(val);return ;}push_down(i);int mid = (L+R)>>1;if(mid<r)update(i<<1|1,l,r,val);if(mid>=l)update(i<<1,l,r,val);push_up(i);
}
long long query(int i,int l,int r)
{int L = tree[i].l;int R = tree[i].r;if(l<=L && R<=r)return tree[i].sum;long long ans = 0;push_down(i);int mid = (L+R)>>1;if(mid<r)ans += query(i<<1|1,l,r);if(mid>=l)ans += query(i<<1,l,r);push_up(i);return ans;
}
int main(void)
{int n,q;while(~scanf("%d %d",&n,&q)){for(int i=1;i<=n;i++)scanf("%d",&a[i]);build(1,1,n);char op[10];while(q--){int x,y;scanf("%s %d %d",op,&x,&y);if(op[0]=='C'){int val;scanf("%d",&val);update(1,x,y,val);}elseprintf("%I64d\n",query(1,x,y));}}return 0;
}