HDU4267 A Simple Problem with Integers 线段树/树状数组
2012长春网络赛A题
Problem Description
Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.
Input
There are a lot of test cases.
The first line contains an integer N. (1 <= N <= 50000)
The second line contains N numbers which are the initial values of A1, A2, ... , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)
The third line contains an integer Q. (1 <= Q <= 50000)
Each of the following Q lines represents an operation.
"1 a b k c" means adding c to each of Ai which satisfies a <= i <= b and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10, -1,000 <= c <= 1,000)
"2 a" means querying the value of Aa. (1 <= a <= N)
Output
For each test case, output several lines to answer all query operations.
Sample Input
4
1 1 1 1
14
2 1
2 2
2 3
2 4
1 2 3 1 2
2 1
2 2
2 3
2 4
1 1 4 2 1
2 1
2 2
2 3
2 4
Sample Output
1
1
1
1
1
3
3
1
2
3
4
1
树状数组：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=50005;
int N,Q;
int val[maxn];
int add[maxn][11][10];
int op,a,b,k,c;
inline int lowbit(int x){ return x & -x;}
inline void _add(int pos, int val, int k, int x){
for(; pos <= N; pos += lowbit(pos)) add[pos][k][x] += val;
}
inline void update(int l, int r, int val, int k, int x){
_add(l, val, k, x); _add(r+1, -val, k, x);
}
int _sum(int pos, int k, int x){
int res = 0;
for(; pos >= 1; pos -= lowbit(pos)) res += add[pos][k][x];
return res;
}
int query(int p)
{
for(int k=1;k<11;k++){
for(int x=0;x<k;x++){
if(p%k==x%k){
int tmp=_sum(p,k,x);
val[p]+=tmp;
update(p,p,-tmp,k,x);
}
}
}
return val[p];
}
int main()
{
freopen("in.txt","r",stdin);
while(~scanf("%d",&N)){
memset(add,0,sizeof(add));
for(int i=1;i<=N;i++) scanf("%d",&val[i]);
scanf("%d",&Q);
while(Q--){
scanf("%d",&op);
if(op==1){
scanf("%d%d%d%d",&a,&b,&k,&c);
update(a,b,c,k,a%k);
}
else{
scanf("%d",&a);
printf("%d\n",query(a));
}
}
}
return 0;
}
线段树：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn=50005;
int N;
struct SegTree
{
int val;
int add[55];
};
SegTree T[maxn*3];
int op,a,b,k,c;
int Q;
int ID[11][11];
void Init()
{
int id=0;
for(int k=1;k<11;k++){
for(int x=0;x<k;x++){
ID[k][x]=id++;
}
}
}
void build(int l,int r,int rt)
{
T[rt].val=0;
memset(T[rt].add,0,sizeof(T[rt].add));
if(l==r){
scanf("%d",&T[rt].val);
return;
}
int m=(l+r)>>1;
build(lson);
build(rson);
}
void push_down(int rt)
{
for(int k=1;k<11;k++){
for(int x=0;x<k;x++){
int id=ID[k][x];
T[rt<<1].add[id]+=T[rt].add[id];
T[rt<<1|1].add[id]+=T[rt].add[id];
T[rt].add[id]=0;
}
}
}
void update(int L,int R,int c,int id,int l,int r,int rt)
{
if(L<=l&&r<=R){
T[rt].add[id]+=c;
return;
}
push_down(rt);
int m=(l+r)>>1;
if(L<=m) update(L,R,c,id,lson);
if(R>m) update(L,R,c,id,rson);
}
int query(int p,int l,int r,int rt)
{
if(l==r){
for(int k=1;k<11;k++){
for(int x=0;x<k;x++){
int id=ID[k][x];
if(l%k==x%k) T[rt].val+=T[rt].add[id];
T[rt].add[id]=0;
}
}
return T[rt].val;
}
push_down(rt);
int m=(l+r)>>1;
if(p<=m) return query(p,lson);
else return query(p,rson);
}
int main()
{
freopen("in.txt","r",stdin);
while(~scanf("%d",&N)){
Init();
build(1,N,1);
scanf("%d",&Q);
while(Q--){
scanf("%d",&op);
if(op==1){
scanf("%d%d%d%d",&a,&b,&k,&c);
int id=ID[k][a%k];
update(a,b,c,id,1,N,1);
}
else{
scanf("%d",&a);
printf("%d\n",query(a,1,N,1));
}
}
}
return 0;
}