标签：des style blog http color os
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 55626 | Accepted: 16755 | |
| 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
Hint
线段树+延迟标记
#include"stdio.h"
#include"string.h"
#define N 100005
int num[N];
struct node
{
int l,r;
__int64 s,a;
}f[N*3];
void creat(int t,int l,int r)
{
f[t].l=l;
f[t].r=r;
f[t].a=0;
if(l==r)
{
f[t].s=num[r];
return ;
}
int temp=t*2,mid=(l+r)/2;
creat(temp,l,mid);
creat(temp+1,mid+1,r);
f[t].s=f[temp].s+f[temp+1].s;
return ;
}
void update(int t,int l,int r,__int64 a)
{
if(f[t].l==l&&f[t].r==r)
{
f[t].a+=a; //每次都是s先增加，a记录下一层的增量
f[t].s+=(f[t].r-f[t].l+1)*a;
return ; //不再向下一层更新
}
int temp=t*2,mid=(f[t].l+f[t].r)/2;
if(f[t].a)
{
f[temp].a+=f[t].a;
f[temp].s+=(f[temp].r-f[temp].l+1)*f[t].a;
f[temp+1].a+=f[t].a;
f[temp+1].s+=(f[temp+1].r-f[temp+1].l+1)*f[t].a;
f[t].a=0; //完成向下一层的加和，进行归零
}
if(r<=mid)
update(temp,l,r,a);
else if(l>mid)
update(temp+1,l,r,a);
else
{
update(temp,l,mid,a);
update(temp+1,mid+1,r,a);
}
f[t].s=f[temp].s+f[temp+1].s;
return ;
}
__int64 find(int t,int l,int r)
{
if(f[t].l==l&&f[t].r==r)
{
return f[t].s;
}
int temp=t*2,mid=(f[t].l+f[t].r)/2;
if(f[t].a) //下一层的更新还没完成
{
f[temp].a+=f[t].a;
f[temp].s+=(f[temp].r-f[temp].l+1)*f[t].a;
f[temp+1].a+=f[t].a;
f[temp+1].s+=(f[temp+1].r-f[temp+1].l+1)*f[t].a;
f[t].a=0;
}
if(r<=mid)
return find(temp,l,r);
else if(l>mid)
return find(temp+1,l,r);
else
return find(temp,l,mid)+find(temp+1,mid+1,r);
}
int main()
{
int i,n,q;
int a,b,c;
char ch;
while(scanf("%d%d",&n,&q)!=-1)
{
for(i=1;i<=n;i++)
scanf("%d",&num[i]);
creat(1,1,n);
while(q--)
{
getchar();
scanf("%c",&ch);
if(ch==‘C‘)
{
scanf("%d%d%d",&a,&b,&c);
update(1,a,b,c);
}
else
{
scanf("%d%d",&a,&b);
printf("%I64d\n",find(1,a,b));
}
}
}
return 0;
}
poj 3468 A Simple Problem with Integers,码迷,mamicode.com
poj 3468 A Simple Problem with Integers
标签：des style blog http color os
原文地址：http://blog.csdn.net/u011721440/article/details/24620215
