敌兵布阵 HDU-1166

版权声明：菜鸟一枚~~ 有想法可在下面评论， 转载标明出处即可。 https://blog.csdn.net/KLFTESPACE/article/details/88674415

参见https://blog.csdn.net/flushhip/article/details/79165701

https://blog.csdn.net/forever_kirito/article/details/78694203 

//树状数组

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>

using namespace std;

#define lowbit(x) (x&(-x)) //计算2^k 

const int Max = 50000+5;
int c[Max];//c[i] = a[i – 2^k + 1] + … + a[i]，k为i在二进制下末尾0的个数

void Update(int x, int y, int MAX)
{
	while(x <= MAX){
		c[x] += y;
		x += lowbit(x);
	}
}

int Sum(int x)
{
	int sum = 0;
	
	while(x > 0){
		sum += c[x];
		x -= lowbit(x);
	}
	return sum;
	
}



void SumSQ(int y, int x)
{
	int sum = 0;
	
	while(y > 0){
		sum += c[y];
		y -= lowbit(y);
	}
	
	while(x > 0){
		sum -= c[x];
		x -= lowbit(x);
	}
	
//	cout << sum << endl;

	printf("%d\n", sum);	
}
int main()
{	
	int T, cnt = 1;
	string s;
	
	//cin >> T;
	scanf("%d", &T);
	
	while(T--){
		int N;
		
	//	cin >> N;
		scanf("%d", &N);
			
		memset(c, 0, sizeof(c));
		
		for(int i=1; i<=N; i++){
			int x;
			
		//	cin >> x;
			scanf("%d", &x);
				
			Update(i, x, N);		
		}
		
//		cout << "Case " << cnt++ << ":" << endl;
        
		printf("Case %d:\n", cnt++);
		
//		while(cin >> s && s != "End"){
		while(cin >> s && s != "End"){

			int x, y;
			
//			cin >> x >> y;
			scanf("%d%d", &x, &y);
						
			if(s == "Add"){
				Update(x, y, N);
			}
			
			else if(s == "Sub"){
				Update(x, -y, N);
			}
			else if(s == "Query"){
				SumSQ(y, x-1);
		//		cout << Sum(y)-Sum(x-1) << endl;
			}
			
		}
		
	}
	return 0;
} 

线段树  //有个比较坑的就是结构体数组的大小问题，，，如果小了的话RE，觉得2倍就可以了，，，结果是TLE。。。。开了3倍或者4倍就过了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

//const int INF = 0x3f3f3f3f;
//const int MOD = 1000000007;

#define LL long long

#define MAXN 50010
 
struct Node
{
    int l, r, val;
}t[MAXN*3];

int ans, c[MAXN];
 
void Build(int x, int y, int num)
{
    t[num].l = x;
    t[num].r = y;
    if(x == y) t[num].val = c[y];
    
    else
    {
        int mid = (x+y) >> 1;
        
		Build(x, mid, num*2);
        Build(mid+1, y, num*2+1);
        
		t[num].val = t[num*2].val + t[num*2+1].val;
    }
}
 
void Add(int x, int y, int num)
{
    if(t[num].l == x && t[num].r == x)
    {
        t[num].val += y;
        return ;
    }
    int mid = (t[num].l + t[num].r) >> 1;
    if(x > mid)
        Add(x, y, num*2+1);
    else
        Add(x, y, num*2);
        
    t[num].val = t[num*2].val + t[num*2+1].val;
}
 
void Query(int x, int y, int num)
{
    if(t[num].l == x && t[num].r == y)
    {
        ans += t[num].val;
        
		return ;
    }
    
    int mid = (t[num].l+t[num].r)>>1;
    if(y <= mid) 
		Query(x, y, num*2);
    else if(x >= mid+1) 
		Query(x, y, num*2+1);
    else {
        Query(x, mid, num*2);
        Query(mid+1, y, num*2+1);
    }
}
 
int main()
{
    int T, n;

    char str[6];
    
	scanf("%d", &T);
    
	for(int i=1; i<=T; i++)
    {
        int x, y;
        scanf("%d", &n);
        
        memset(c, 0, sizeof(c)); 
        
		for(int j=1; j<=n; j++)
            scanf("%d", &c[j]);
        
		Build(1, n, 1);
        
		printf("Case %d:\n", i);
		
        while(~scanf("%s", str))
        {
            
			if(strcmp(str, "End")==0) 
				break;
            else if(strcmp(str, "Add")==0)
            {
                scanf("%d%d", &x, &y);
                Add(x, y, 1);
            }
            else if(strcmp(str, "Sub")==0)
            {
                scanf("%d%d", &x, &y);
                Add(x, -y, 1);
            }
            else if(strcmp(str, "Query") == 0)
            {
                scanf("%d%d", &x, &y);
                
				ans = 0;
                
                Query(x, y, 1);
                
                printf("%d\n", ans);
            }
        }
    }
    return 0;
}