标签：des style color 使用 os io strong for
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 61636 | Accepted: 18840 | |
| 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
题解及代码：
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <set>
#include <map>
#include <queue>
#include <string>
#define maxn 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ALL %I64d
using namespace std;
typedef __int64 ll;
struct segment
{
int l,r;
ll value;
ll nv;
} son[maxn<<2];
void PushUp(int rt)
{
son[rt].value=son[rt<<1].value+son[rt<<1|1].value;
}
void Build(int l,int r,int rt)
{
son[rt].l=l;
son[rt].r=r;
son[rt].nv=0;
if(l==r)
{
scanf("%I64d",&son[rt].value);
return;
}
int m=(l+r)/2;
Build(lson);
Build(rson);
PushUp(rt);
}
void Update_n(ll w,int l,int r,int rt)
{
if(son[rt].l==l&&son[rt].r==r)
{
son[rt].value+=w*(r-l+1);
son[rt].nv+=w;
return;
}
if(son[rt].nv)
{
son[rt<<1].nv+=son[rt].nv;
son[rt<<1|1].nv+=son[rt].nv;
son[rt<<1].value+=(son[rt<<1].r-son[rt<<1].l+1)*son[rt].nv;
son[rt<<1|1].value+=(son[rt<<1|1].r-son[rt<<1|1].l+1)*son[rt].nv;
son[rt].nv=0;
}
int m=(son[rt].l+son[rt].r)/2;
if(r<=m)
Update_n(w,l,r,rt<<1);
else if(l>m)
Update_n(w,l,r,rt<<1|1);
else
{
Update_n(w,lson);
Update_n(w,rson);
}
PushUp(rt);
}
ll Query(int l,int r,int rt)
{
if(son[rt].l==l&&son[rt].r==r)
{
return son[rt].value;
}
if(son[rt].nv)
{
son[rt<<1].nv+=son[rt].nv;
son[rt<<1|1].nv+=son[rt].nv;
son[rt<<1].value+=(son[rt<<1].r-son[rt<<1].l+1)*son[rt].nv;
son[rt<<1|1].value+=(son[rt<<1|1].r-son[rt<<1|1].l+1)*son[rt].nv;
son[rt].nv=0;
}
ll ret=0;
int m=(son[rt].l+son[rt].r)/2;
if(r<=m)
ret=Query(l,r,rt<<1);
else if(l>m)
ret=Query(l,r,rt<<1|1);
else
{
ret=Query(lson);
ret+=Query(rson);
}
//PushUp(rt);
return ret;
}
int main()
{
int n,m,l,r;
ll w;
char s[4];
while(scanf("%d%d",&n,&m)!=EOF)
{
Build(1,n,1);
for(int i=1; i<=m; i++)
{
scanf("%s",s);
if(s[0]=='C')
{
scanf("%d%d%I64d",&l,&r,&w);
Update_n(w,l,r,1);
}
else
{
scanf("%d%d",&l,&r);
printf("%I64d\n",Query(l,r,1));
}
}
}
return 0;
}
/*
作为初学者，写起程序也总是会出现一些小毛病，都是对于线段树的理解不深造成的。
说一下这道题目需要注意的地方，那就是无论是每个节点的值还是增量，都要使用long long，
否则会错掉。
经过这段时间的学习，弄懂了线段树的好多地方，现在感觉做题也没那么费力了，都是简单题目，
对于初学者来说还是比较好的，欢迎大家一起讨论学习。
*/
poj 3468 A Simple Problem with Integers（线段树）,布布扣,bubuko.com
poj 3468 A Simple Problem with Integers（线段树）
标签：des style color 使用 os io strong for
原文地址：http://blog.csdn.net/knight_kaka/article/details/38581923
