C. Ehab and a 2-operation tasktime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputYou’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?InputThe 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.OutputOn 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 ii or xx don’t satisfy the limitations above, or you use more than n+1operations, you’ll get wrong answer verdict.ExamplesinputCopy3
1 2 3
outputCopy0inputCopy3
7 6 3
outputCopy2
1 1 1
2 2 4
NoteIn 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].

#include<bits/stdc++.h>
using namespace std;
int num=1e6,i;
int main()
{
    int a;
    cin>>a;
    cout<<a+1<<endl;
    //cout<<1<<" "<<a<<" "<<1<<endl;
    cout<<2<<" "<<a<<" "<<1<<endl;
    cout<<1<<" "<<a<<" "<<num<<endl;
    while(--a)
    {
        printf("2 %d %d\n",++i,--num);
    }
    return 0;
}

 

這道題，我沒想到1e6會用到，還有n+1，因為是找出存在的解，所以找到一般需要技巧，驗證解是否存在，我的常規想法往往是找出最優解，所以會陷入誤區，所以存在性問題要會找答案。由於%1所有數都會為0，很特殊，所以直接轉化，1e6很大的數也是有 特徵的，直接各個加上，模上不超過2000的數，沒有影響，又是不超過n+1步，所以直接搞成遞增就行了
哇，一看程式碼，瞬間覺得自己傻叉了，還硬算，如此簡潔的程式碼，還是要充分利用條件的