Codeforces Round #179 (Div. 1)-A. Greg and Array

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Greg has an array a = a1, a2, ..., an and
m operations. Each operation looks as:
li,
ri,
di,
(1 ≤ li ≤ ri ≤ n). To apply operation
i to the array means to increase all array elements with numbers
li, li + 1, ..., ri by value
di.

Greg wrote down k queries on a piece of paper. Each query has the following form:
xi,
yi,
(1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers
xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input
The first line contains integers n,
m, k
(1 ≤ n, m, k ≤ 105). The second line contains
n integers: a1, a2, ..., an
(0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number
i is written as three integers:
li,
ri,
di,
(1 ≤ li ≤ ri ≤ n),
(0 ≤ di ≤ 105).

Next k lines contain the queries, the query number
i is written as two integers:
xi,
yi,
(1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output
On a single line print n integers
a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in
C++. It is preferred to use the
cin, cout streams of the
%I64d specifier.

Sample test(s)

Input

3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3

Output

9 18 17

Input

1 1 1
1
1 1 1
1 1

Output

2

Input

4 3 6
1 2 3 4
1 2 1
2 3 23 4 4
1 21 3
2 3
1 21 3
2 3

Output

5 18 31 20

题意：

从数L到R加上D。

从操作X到Y 对数组进行操作。

思路：

数组vis 对操作数进行记录，数组q 对要修改的数进行记录。数组区间下标中左端点++，（右端点+1）- -，将数组进行一次循环即可记录所有数的情况。

CODE：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <map>
typedef long long ll;
const int inf=0xffffff;
const int M=100005;
using namespace std;

struct node
{
int l,r,d;
}nn[M];
ll num[M],q[M],vis[M];

int main()
{
//freopen("in.in","r",stdin);
int n,m,k;
while(~scanf("%d %d %d",&n,&m,&k))
{
memset(vis,0,sizeof(vis));
memset(num,0,sizeof(num));
memset(q,0,sizeof(q));
for(int i=1;i<=n;i++)
scanf("%lld",&num[i]);
for(int i=1;i<=m;i++)
{
scanf("%d %d %d",&nn[i].l,&nn[i].r,&nn[i].d);
nn[i].r++;
}
int x,y;
for(int i=0;i<k;i++)
{
scanf("%d%d",&x,&y);
vis[x]++;
vis[y+1]--;
}
for(int i=1;i<=m;i++)
{
vis[i] += vis[i-1];
}

for(int i=1;i<=m;i++)
{
q[nn[i].l] += vis[i]*nn[i].d;
q[nn[i].r] -= vis[i]*nn[i].d;
}
for(int i=1;i<=n;i++)
{
q[i] += q[i-1];
}
printf("%I64d",num[1]+q[1]);
for(int i=2;i<=n;i++)
printf(" %I64d",num[i]+q[i]);
printf("\n");
}

return 0;
}