A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 105742 | Accepted: 33031 | |
| Case Time Limit: 2000MS | ||
Description
You have N integers, A1, A2, ... ,
AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1,
A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa,
Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... ,
Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
思路：线段树区间跟新 + lazy-target标记
（注意数据范围，long long）
代码：
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdio>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int maxn=100005;
long long sum[maxn<<2];
long long add[maxn<<2];
void pushup(int rt) {
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt, int len) {
if(add[rt]) {
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=(len-(len>>1))*add[rt];
sum[rt<<1|1]+=(len>>1)*add[rt];
add[rt]=0;
}
}
void build(int l, int r, int rt) {
add[rt]=0;
if(l==r) {
scanf("%lld",&sum[rt]);
return;
}
int mid=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(int L, int R, int val, int l, int r, int rt) {
if(L<=l&&r<=R) {
sum[rt]+=(r-l+1)*val;
add[rt]+=val;
return;
}
pushdown(rt,r-l+1);
int mid=(l+r)>>1;
if(L<=mid) update(L,R,val,lson);
if(R>mid) update(L,R,val,rson);
pushup(rt);
}
long long query(int L, int R, int l, int r, int rt) {
if(L<=l&&r<=R) {
return sum[rt];
}
pushdown(rt,r-l+1);
int mid=(l+r)>>1;
long long cnt=0;
if(L<=mid) cnt+=query(L,R,lson);
if(R>mid) cnt+=query(L,R,rson);
return cnt;
}
int main() {
int n,q;
while(~scanf("%d%d",&n,&q)) {
build(1,n,1);
char s[10];
for(int i=1;i<=q;i++) {
scanf("%s",s);
if(s[0]=='Q') {
int L,R;long long result;scanf("%d%d",&L,&R);
result=query(L,R,1,n,1);
printf("%lld\n",result);
} else if(s[0]=='C') {
int L,R,val;scanf("%d%d%d",&L,&R,&val);
update(L,R,val,1,n,1);
}
}
}
return 0;
}