Robin Hood likes to loot rich people since he helps the poor people with this money. Instead of keeping all the money together he does another trick. He keeps n sacks where he keeps this money. The sacks are numbered from 0 to n-1.
Now each time he can he can do one of the three tasks.
1) Give all the money of the ith sack to the poor, leaving the sack empty.
2) Add new amount (given in input) in the ith sack.
3) Find the total amount of money from ith sack to jth sack.
Since he is not a programmer, he seeks your help.
Input starts with an integer T (≤ 5), denoting the number of test cases.
Each case contains two integers n (1 ≤ n ≤ 105) and q (1 ≤ q ≤ 50000). The next line contains n space separated integers in the range [0, 1000]. The ith integer
denotes the initial amount of money in the ith sack (0 ≤ i < n).
Each of the next q lines contains a task in one of the following form:
1 i Give all the money of the ith(0 ≤ i < n) sack to the poor.
2 i v Add money v (1 ≤ v ≤ 1000) to the ith(0 ≤ i < n) sack.
3 i j Find the total amount of money from ith sack to jth sack (0 ≤ i ≤ j < n).
For each test case, print the case number first. If the query type is 1,
then print the amount of money given to the poor. If the query type is 3,
print the total amount from ith to jth sack.
1
5 6
3 2 1 4 5
1 4
2 3 4
3 0 3
1 2
3 0 4
1 1
Case 1:
5
14
1
13
2
题意：
输入T组测试数据：
N个数据，Q个操作；
1，i 将编号为i的数变成0，并且输出被改变的点的数值；
2，i，v 将编号为i的数加v;
3，i，j 输出i~j之间的总和；
思路：
线段树或者树状数组直接就over了；
这个题和hdoj 1166 点击打开链接思路基本一样，单点更新
注意点：
这是平时是训练的一道题目，我有点遗憾，就在最后1秒钟想到哪儿错了
。。。。。。。
1 ,i 不是直接输出，而是通过query或者getsum求，我把他当成直接输出了
线段树:
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
#define lson l,mid,v<<1
#define rson mid+1,r,v<<1|1
typedef long long int LL;
const int MAX=100000;
int sum[4*MAX+20];
int num[MAX+5];
void build (int l,int r,int v)
{
sum[v]=0;
if(l==r)
{
sum[v]=num[l];
return ;
}
int mid=(l+r)>>1;
build(lson);
build(rson);
sum[v]=sum[2*v]+sum[2*v+1] ;
}
void update(int l,int r,int v,int pos,int x)
{
if(l==r)
{
sum[v]+=x;
return ;
}
int mid=(l+r)>>1;
if(pos<=mid)
update(lson,pos,x);
else
update(rson,pos,x);
sum[v]=sum[2*v]+sum[2*v+1] ;
}
int query(int l,int r,int v,int lpos,int rpos)
{
if(lpos<=l&&r<=rpos)
return sum[v];
int mid=(l+r)>>1;
if(rpos<=mid)
return query(lson,lpos,rpos);
else if(lpos>mid)
return query(rson,lpos,rpos);
else
return query(lson,lpos,mid)+query(rson,mid+1,rpos);
}
int main()
{
int T,ca=1;
scanf("%d",&T);
while(T--)
{
int n,m;
scanf("%d%d",&n,&m);
memset(num,0,sizeof(num));
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
build(1,n,1);
printf("Case %d:\n",ca++);
while(m--)
{
int opr;
int a=0,b=0,c=0;
scanf("%d",&opr);
if(opr==1)
{
scanf("%d",&a);
a++;
printf("%d\n",query(1,n,1,a,a));//这儿坑爹
update(1,n,1,a,-query(1,n,1,a,a));//注意
}
if(opr==2)
{
scanf("%d%d",&a,&b);
a++;
update(1,n,1,a,b);
}
if(opr==3)
{
scanf("%d%d",&a,&b);
a++;b++;
printf("%d\n",query(1,n,1,a,b));
}
}
}
return 0;
}
树状数组：
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long int LL;
LL num[500001];
int v[500001];
int n;
int lowbit(int x)
{
return x&(-x);
}
LL getsum(int i)
{
int sum=0;
while(i>0)
{
sum+=num[i];
i-=lowbit(i);
}
return sum;
}
void update(int i,int m)
{
while(i<=n)
{
num[i]+=m;
i+=lowbit(i);
}
}
int main()
{
int T,ca=1;
scanf("%d",&T);
while(T--)
{
int m;
scanf("%d%d",&n,&m);
int i,x;
memset(num,0,sizeof(num));
for(i=1;i<=n;i++)
{
scanf("%d",&v[i]);
update(i,v[i]);
}
printf("Case %d:\n",ca++);
while(m--)
{
int opr;
scanf("%d",&opr);
int l,r;
if(opr==1)
{
scanf("%d",&l);
l++;
printf("%d\n",getsum(l)-getsum(l-1));//注意
update(l,getsum(l-1)-getsum(l)) ;//注意
}
if(opr==2)
{
scanf("%d%d",&l,&r);
l++,r;
update(l,r);
}
if(opr==3)
{
scanf("%d%d",&l,&r);
l++;r++;
printf("%d\n",getsum(r)-getsum(l-1));
}
}
}
return 0;
}
版权声明：本文为博主原创文章，未经博主允许不得转载。