A.
Input
The only line contains the integer xx (1≤x≤100)(1≤x≤100).
Output
You should output two integers aa and bb, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).
Examples
Input
10
Output
6 3Input
1
Output
-1解题报告：
就是给你一个x，让你构造一个a和b，满足和x的这仨关系就行了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int main()
{
int x,a,b;
cin>>x;
int flag = 0;
for(int i = 1; i<=x; i++) {
for(int j = i; j<=x; j+=i) {
if(i*j>x && j/i < x) {
flag = 1;
a=j;b=i;break;
}
}
if(flag) break;
}
if(flag) printf("%d %d\n",a,b);
else puts("-1");
return 0 ;
}B.
给你n个数的数组，给一个k。让你重复执行k次下列操作：每次选中序列中不为0的最小的数并输出，然后把序列中所有非零的数都减去这个选中的数。如果序列已经都是0了，那就输出0.
Examples
Input
3 5
1 2 3
Output
1
1
1
0
0
Input
4 2
10 3 5 3
Output
3
2Note
In the first sample:
In the first step: the array is [1,2,3][1,2,3], so the minimum non-zero element is 1.
In the second step: the array is [0,1,2][0,1,2], so the minimum non-zero element is 1.
In the third step: the array is [0,0,1][0,0,1], so the minimum non-zero element is 1.
In the fourth and fifth step: the array is [0,0,0][0,0,0], so we printed 0.
In the second sample:
In the first step: the array is [10,3,5,3][10,3,5,3], so the minimum non-zero element is 3.
In the second step: the array is [7,0,2,0][7,0,2,0], so the minimum non-zero element is 2.
解题报告：
先别看哪个Node，，有的时候会误导你的思路。。其实这题就是将减的数给可持久化一下、、因为你想啊你总不能减一个数，就将序列中的数都变化一下吧，，所以解决方法就是用一个sub来动态记录变化的数的值。为啥可以这样呢？因为你虽然要变化整个数组中的值，，但是其实你就算不变化，，也不会影响下次你取数组中的最小值（因为大家都减同一个数，相对的大小关系肯定没变啊）所以就用一个变量记录就行了。
AC代码1：（排序版）
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a[MAX];
int main() {
int n,k;
cin>>n>>k;
for(int i=1; i<=n; i++) scanf("%lld",a+i);
ll sub=0;
sort(a+1,a+n+1);
for(int i=1,j=1;j<=k;i++,j++) {
if(i>n) {
printf("0\n");
}
else {
while(1) {
if(a[i]!=sub) break;
i++;
if(i > n) {
printf("0\n");break;
}
}
if(i<=n) {
a[i]-=sub;
printf("%lld\n",a[i]);
sub+=a[i];
}
}
}
return 0;
}或者用优先队列也可以写:
AC代码2：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f
using namespace std;
priority_queue<int, vector<int>, greater<int> > pq;
int n,k;
int main() {
cin>>n>>k;
for(int i = 1,tmp; i <= n; i ++) {
scanf("%d",&tmp);
if(tmp) pq.push(tmp);
}
int sub = 0;
while(k--) {
if(!pq.empty()) {
int cur = pq.top();
while(cur - sub == 0 && !pq.empty()) {
pq.pop();
cur=pq.top();
}
printf("%d\n",cur-sub);
sub += (cur-sub);
}
else puts("0");
}
return 0;
}
不过更适合我的方式其实是先判断一堆直到只能变成0。然后再看看剩余的次数是否大于0，如果大于0再输出k次的0就行了、
C.
不想写题意了。。**构造一眼题
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 ii or xx don't satisfy the limitations above, or you use more than n+1n+1 operations, you'll get wrong answer verdict.
Examples
Input
3 1 2 3
Output
0
Input
3 7 6 3
Output
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].
好吧我还是写一下题意。。就是说给你n个数，给你两种操作：1.任取一个x，让区间内所有的数都加上这个x。2.任取一个x，让区间内所有的数都取模这个x。(其中x<=1e6)。使得这个序列变成一个严格递增序列，让你输出每次操作是什么，和怎么操作、、
解题报告：
1分钟出标算，，10分钟WA两发，，刚开始这个大数去了1e7，，（没读题），，后来WA1，改成1e6，感觉该没问题了吧，WA2，，发现后面在取模的时候去的x会超过1e6，，再改成5e5，AC。。
其实这题上来就取个5000左右就行了啊。。
不难构造，，题意都提示的很明显了，，n+1次操作，那肯定猜一个n次一个1次呗，，然后不难想到加一个大数，然后每次都取模，让他构造成一个1~n的序列就完事了。
（白白丢掉100分啊，，，心疼）（另外，应该输出a[i]-i，，别写成了a[i]-1，，，）
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
ll a[MAX];
int main()
{
int n;
cin>>n;
for(int i = 1; i<=n; i++) scanf("%lld",a+i);
printf("%d\n",n+1);
printf("1 %d 500000\n",n);
for(int i = 1; i<=n; i++) a[i] += 500000;
for(int i = 1; i<=n; i++) {
printf("2 %d %lld\n",i,a[i]-i);
}
return 0 ;
}D.
未补。抽空学学交互题吧、、