Codeforces Round #179 (Div. 2) C. Greg and Array
Greg and Array
time limit per test 2 seconds
memory limit per test 56 megabytes
input standard input
output standard output
Greg has an array a = a1, a2, ..., an andm operations. Each operation looks as: li,ri,di,(1 ≤ li ≤ ri ≤ n). To apply operationi to the array means to increase all array elements with numbersli, li + 1, ..., ri by valuedi.
Greg wrote down k queries on a piece of 易做图. Each query has the following form:xi,yi,(1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbersxi, 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 containsn integers:a1, a2, ..., an(0 ≤ ai ≤ 105) — the initial array.
Next m lines contain operations, the operation numberi is written as three integers:li,ri,di,(1 ≤ li ≤ ri ≤ n),(0 ≤ di ≤ 105).
Next k lines contain the queries, the query numberi 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 inC++. It is preferred to use thecin, 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 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20
这题真是太遗憾了，有一个地方下标应该从1开始写成了0，结果浪费了5分多钟的时间找bug，结果比赛就结束了，遗憾啊。 考察点：树状数组 思路：建立两个树状数组，分别统计操作实行的次数和每次操作数值的变化
[cpp]
#include <stdio.h>
#include <string.h>
#include <math.h>
__int64 tree1[110000],tree2[110000];
struct num {
int l,r,val; }a[110000];
__int64 b[1000000],sum[110000];
__int64 n,m; int main()
{ void build(__int64 k,__int64 val);
void build2(__int64 k,__int64 val);
__int64 search(__int64 k);
__int64 search2(__int64 k);
__int64 i,j,s,t;
__int64 x,y,k;
while(scanf("%I64d %I64d %I64d",&n,&m,&k)!=EOF)
{ for(i=1;i<=n;i++)
{ scanf("%I64d",&b[i]); }
for(i=1;i<=m;i++)
{ scanf("%d %d %d",&a[i].l,&a[i].r,&a[i].val);
} memset(tree2,0,sizeof(tree2));
memset(tree1,0,sizeof(tree1)); for(i=1;i<=k;i++)
{ scanf("%I64d %I64d",&x,&y);
build(y,1);
build(x-1,-1); }
memset(sum,0,sizeof(sum));
补充：软件开发 , C++ ,
