HDU 1166

http://acm.hdu.edu.cn/showproblem.php?pid=1166

分析：

这里给出树状数组和线段树两种方法   不要介意两个不同的头文件

AC代码（树状数组）：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
#define lson 2*i
#define rson 2*i+1 
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for (i=x;i<=y;i++)
#define DOWN(i,x,y) for (i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL longlong 
#define N 1000005
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)



int c[N];
int t,n;
void add(int k,int num){
    W(k<=n){
        c[k]+=num;
        if(c[k]<0)
            c[k]=0;
        k+=lowbit(k);
    }
} 
int read(int k){
    int sum=0;
    W(k){
        sum+=c[k];
        k-=lowbit(k);
    }
    return sum;
}
int main (){
    int i;
    scanf ("%d",&t);
    int count=1;
    W(t--){
        //freopen("data.txt","r",stdin);
        MEM(c,0);
        scanf("%d",&n);
        UP(i,1,n){
            int temp;
            scanf ("%d",&temp);
            add(i,temp);
        }
        printf ("Case %d:\n",count++);
        char str[20];
        int x,y;
        W(scanf("%s",str)&&str[0]!='E'){
            scanf ("%d%d",&x,&y);
            if (str[0]=='A')
                add(x,y);
            else if(str[0]=='S')
                add(x,-y);
            else if(str[0]=='Q')
                printf ("%d\n",read(y)-read(x-1));         
        }
    }
    return 0;
} 

AC代码 （线段树）：

#include <iostream>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include<list>
#include <bitset>
#include <climits>
#include <algorithm>
#define gcd(a,b) __gcd(a,b)
#define FIN	freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w",stdout)
typedef long long LL;
const LL mod=1e9+7;
const int INF=0x3f3f3f3f;
const int MAX=1e6+5;
const double PI=acos(-1.0);
using namespace std;
LL a[MAX];
struct Tree{
	int l,r;
	LL val;
	LL add; //      标记 
}T[MAX];
void build(int rt,int l,int r){//     建树  
	T[rt].add=0;
	T[rt].l=l;
	T[rt].r=r;
	if (l==r) {
		T[rt].val=a[l];
		return ; 
	} 
	int mid=(l+r)/2;
	build(rt*2,l,mid);
	build(rt*2+1,mid+1,r);
	T[rt].val=T[rt*2].val+T[rt*2+1].val;
} 
void pushDown(int rt,int l,int r){//        标记下推 
	if (T[rt].add!=0){
		T[rt*2].add+=T[rt].add;
		T[rt*2+1].add+=T[rt].add;
		T[rt*2].val+=T[rt].add*l;
		T[rt*2+1].val+=T[rt].add*r; 
		T[rt].add=0;
	}
}
void update(int rt,int l,int r,int e){//      修改区间
	if (T[rt].l>=l&&T[rt].r<=r){
		T[rt].add+=e;
		T[rt].val+=e*(T[rt].r-T[rt].l+1);
		return ;
	}
	int mid=(T[rt].l+T[rt].r)/2;
	pushDown(rt,mid-T[rt].l+1,T[rt].r-mid);
	if (l<=mid) update(rt*2,l,r,e);
	if (r>mid) update(rt*2+1,l,r,e);
	T[rt].val=T[rt*2].val+T[rt*2+1].val;
} 
LL Query (int rt,int Ql,int Qr){//    查询 
	if (T[rt].l>=Ql&&T[rt].r<=Qr) return T[rt].val;
	int mid=(T[rt].l+T[rt].r)/2;
	pushDown(rt,mid-T[rt].l+1,T[rt].r-mid);
	LL ans=0;
	if (Ql<=mid) ans+=Query(rt*2,Ql,Qr);
	if (Qr>mid) ans+=Query(rt*2+1,Ql,Qr);
	return ans;
}
int main (){
	int cc=1; 
	int t;
	//FIN;
	scanf ("%d",&t);
	while(t--){
		printf ("Case %d:\n",cc++);
		int n;
		scanf ("%d",&n);
		for (int i=1;i<=n;i++) scanf ("%lld",&a[i]);
		build(1,1,n);
		int m;
		char st[20];
			while (scanf ("%s",st)&&st[0]!='E'){
				if (st[0]=='A'){
					int s,v;
					scanf ("%d%d",&s,&v);
					update(1,s,s,v);
				}
				else if(st[0]=='S'){
					int s,v;
					scanf ("%d%d",&s,&v);
					update(1,s,s,-v);
				}
				else{
					int s,e;
					scanf ("%d%d",&s,&e);
					printf ("%lld\n",Query(1,s,e));
				}
			}
			
	} 
	return 0;
}