hdu1166-敌兵布阵
题意:
读入一个数组,完成单点加,单点减,和区间求和三种操作。
解法:
明摆着的一颗线段树,SB题。
CODE:
#pragma GCC optimize(2)
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL long long
#define N 100010
struct Tree {
int lson,rson;
int sum,tag;
}tree[N * 2];
int n,cnt = 1,a[N],ans,T,test;
char s[N];
inline int read(){
int x = 0,f = 1;
char ch = getchar();
while(ch<'0'||ch>'9') {if(ch == '-') f = -1; ch = getchar();}
while(ch>='0'&&ch<='9') {x = (x << 1) + (x << 3) + (ch ^ 48);ch = getchar();}
return x * f;
}
inline void pushup(int x) {
tree[x].sum = tree[tree[x].lson].sum + tree[tree[x].rson].sum;
}
inline void pushdown(int x,int l,int r) {
int mid = (l + r) >> 1;
tree[tree[x].lson].sum += (mid - l + 1) * tree[x].tag;
tree[tree[x].lson].tag += tree[x].tag;
tree[tree[x].rson].sum += (r - mid) * tree[x].tag;
tree[tree[x].rson].tag += tree[x].tag;
tree[x].tag = 0;
}
void build(int x,int l,int r) {
if(l == r) {
tree[x].sum = a[l];
return;
}
int mid = (l + r) >> 1;
tree[x].lson = ++cnt;
build(tree[x].lson,l,mid);
tree[x].rson = ++cnt;
build(tree[x].rson,mid+1,r);
pushup(x);
}
void update(int x,int l,int r,int k,int v) {
if(l == r) {
tree[x].sum += v;
return;
}
int mid = (l + r) >> 1;
if(k <= mid) update(tree[x].lson,l,mid,k,v);
else update(tree[x].rson,mid+1,r,k,v);
pushup(x);
}
int query(int x,int l,int r,int ll,int rr) {
if(l == ll && r == rr) return tree[x].sum;
int mid = (l + r) >> 1;
pushdown(x,l,r);
if(rr <= mid) return query(tree[x].lson,l,mid,ll,rr);
else if(ll > mid) return query(tree[x].rson,mid+1,r,ll,rr);
else return query(tree[x].lson,l,mid,ll,mid) + query(tree[x].rson,mid+1,r,mid+1,rr);
}
int main() {
T = read();
while(T--) {
test++;
n = read();
for(int i = 1 ; i <= n ; i++)
a[i] = read();
build(1,1,n);
printf("Case %d:\n",test);
while(true) {
scanf("%s",s+1);
if(s[1] == 'E') break;
if(s[1] == 'A') {
int i = read() , j = read();
update(1,1,n,i,j);
}
if(s[1] == 'S') {
int i = read() , j = read();
update(1,1,n,i,-j);
}
if(s[1] == 'Q') {
int i = read() , j = read();
printf("%d\n",query(1,1,n,i,j));
}
}
}
//system("pause");
return 0;
}