C. Ehab and a 2-operation task (cf)
| Codeforces Round #525 (Div. 2) |
|---|
| Finished
|
C. Ehab and a 2-operation task
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
You're given an array aa of length nn. You can perform the following operations on it:
- choose an index ii (1≤i≤n)(1≤i≤n), an integer xx (0≤x≤106)(0≤x≤106), and replace ajaj with aj+xaj+x for all (1≤j≤i)(1≤j≤i), which means add xx to all the elements in the prefix ending at ii.
- choose an index ii (1≤i≤n)(1≤i≤n), an integer xx (1≤x≤106)(1≤x≤106), and replace ajaj with aj%xaj%x for all (1≤j≤i)(1≤j≤i), which means replace every element in the prefix ending at ii with the remainder after dividing it by xx.
Can you make the array strictly increasing in no more than n+1n+1 operations?
Input
The first line contains an integer nn (1≤n≤2000)(1≤n≤2000), the number of elements in the array aa.
The second line contains nn space-separated integers a1a1, a2a2, ……, anan (0≤ai≤105)(0≤ai≤105), the elements of the array aa.
Output
On the first line, print the number of operations you wish to perform. On the next lines, you should print the operations.
To print an adding operation, use the format "11 ii xx"; to print a modding operation, use the format "22 ii xx". If iior xx don't satisfy the limitations above, or you use more than n+1n+1 operations, you'll get wrong answerverdict.
Examples
input
Copy
3 1 2 3
output
Copy
0
input
Copy
3 7 6 3
output
Copy
2 1 1 1 2 2 4
Note
In the first sample, the array is already increasing so we don't need any operations.
In the second sample:
In the first step: the array becomes [8,6,3][8,6,3].
In the second step: the array becomes [0,2,3][0,2,3].
题意：用最多n+1次操作把数组变成严格递增的
#include
using namespace std;
#define ll long long
const int N = 1e6 + 7;
int a[N];
int b[N], c[N], d[N], cnt = 0;
int main() {
int n;
scanf ("%d", &n);
for (int i = 1; i <= n; ++i) {
scanf ("%d", &a[i]);
}
printf ("%d\n", n+1);
for (int i = n; i >= 1; --i) {
int x = (i-1) - (a[i]%n) + n;
printf ("1 %d %d\n", i, x);
for (int j = 1; j <= i; ++j) {
a[j] += x;
}
}
printf ("2 %d %d\n", n, n);
/*puts("-------------");
for (int i = 1; i <= n; ++i) {
a[i]%=n;
printf ("%d ", a[i]);
}*/
return 0;
}