This article is an English translation of the NOIP Round 1 (the Qualification Round) problems.

Here are some backgrounds about it.

- matthew99 has written an Introduction to Chinese OI contests, but he ignored this for some reason; maybe this exam is too easy for him in his province.
- It is literally a "preliminary contest".
- Basically, this is a qualification round for NOIP.
- This is a written examination.
- The exam lasts for 2 hours.
- The total score is 100. You may need to get ~80 scores (in provinces like Zhejiang) or ~20 scores (in provinces like Inner Mongolia) to pass.
- This translation may be not 100% exact, but you can feel the flavor of the NOIPR1.

In part IV and V, the translator modified and formatted the source codes with `clang-format`

.

### Part I: Single Choice Problems

2 score for each problem.

In the following four numbers in different positional notation, which one is different from the other three? A. (269)16 B. (617)10 C. (1151)8 D. (1001101011)2

Which of the following programming language is run by a interpreter? A. C B. C++ C. Pascal D. Python

When did Chinese Computer Federation founded the NOI competitions? (it was called "National High School Students' Computer Programming Contest") A. 1983 B. 1984 C. 1985 D. 1986

Let the depth of a single node to be 0. What is the number of nodes of a full k-ary tree of depth h? A. (

*k*^{h + 1}- 1) / (*k*- 1) B.*k*^{h - 1}C.*k*^{h}D. (*k*^{h - 1}) / (*k*- 1)There is an algorithm whose time-cost function is T(n) = T(n — 1) + n, T(0)=1. What is its time complexity? A. B. C.

*O*(*n*) D.*O*(*n*^{2})What is the prefix form (Polish notation) of the expression

`a*d-b*c`

A.`a d * b c * -`

B.`- * a d * b c`

C.`a * d - b * c`

D.`- * * a d b c`

Choose two points A and B randomly on a segment of length 1. What's the expected length of |

*AB*| ? A. 1 / 2 B. 1 / 3 C. 2 / 3 D. 3 / 5Catalan Numbers is

*C*_{n}= (2*n*)! / (*n*+ 1)! /*n*! . Which of the following is wrong? A.*C*_{n}= the number of different binary trees of n+1 nodes B.*C*_{n}= the number of legal bracket sequence of length 2n C.*C*_{n}= the number of possible "pop sequence" if we push 1,2,...,n into a stack D.*C*_{n}= the number of ways to triangulate a convex (n+2)-gonAssume there is a machine with a button. Whenever you push that button, it will randomly give you either a red ball or a blue ball with 1/2 probability each. Infinitively many people play with this machine. Each of them push that button until a blue ball is given. Then they will put all their balls into a big box. What is the ratio of the two colors of the balls in the box? A. 1 : 2 B. 2 : 1 C. 1 : 3 D. 1 : 1

You need to write a code to compute the number of 1's in a binary form of a number n. Here is the code:

`int CountBit(int x) { int ret = 0; while (x) { ret++; ________; } return ret; }`

What should you fill in the blank? A.

`x >>= 1`

B.`x &= x - 1`

C.`x |= x >> 1`

D.`x <<= 1`

### Part II: Multiple Choice Problems

There are several choices in each problem. You must find them all to get 2 score for each problem in this part.

- What can you take into the exam room in NOIP Round 1? A. Pen or Pencil B. Eraser C. Mobile phone (turned off) D. Draft paper
- 2-3 trees are a special kind of tree. They satisfy: (1) all non-leaf nodes have 2 or 3 children; (2) all leaf has the same depth. If a 2-3 tree have 10 leaves, how many non-leaf nodes may it have? A. 5 B. 6 C. 7 D. 8
- Which of the following statements is true? A. if there is no negative cycle but negative edges in a graph, Dijkstra's algorithm may not calculate the single source shortest paths B. if there is no negative edge, calling Dijkstra's algorithm multiple times may calculate the shortest paths between any pair of nodes C. if there is negative cycles, you can call Dijkstra's algorithm one time to calculate single source shortest paths D. when there is no negative edge, calling Dijkstra's algorithm once cannot be used to calculate the shortest paths between any pair of nodes
- Which of the following property does a tree have? A. there is no cycle B. there is only one simple path between any pair of nodes C. have and only have one simple cycle D. the number of edges equals to the number of nodes minus one
- Which of the following statements about the Turing Award is true? A. it is set up by Institute of Electrical and Electronics Engineers (IEEE) B. the only Chinese who has got the Award is Yao Qizhi C. its name is from the pioneer of computer science, the British scientist Alan Mathison Turing D. it's the most famous and most honored award in computer field. It was honored as the "Nobel Prize in computer field".

### Part III: Problem Solving

5 score for each task.

- There are four friends. let's call them Jia, Yi, Bing and Ding. They are deciding whether they will go out for a hike. (1) if it rains and Yi doesn't go out, then Jia will not go out; (2) if Yi goes out, then Ding will go out; (3) if Bing goes out, Ding must not go out; (4) if Ding doesn't go out and Jia doesn't go out, then Bing won't go out. Now you can see Bing is going out. So: Does Jia go out? Does Yi go out? Does Ding go out? Is it raining? The last problem is 2 scores; the other 1 score.
- How many pairs of solution does the equation
`a*b=(a and b)*(a or b)`

have when`a,b`

are integers in range [0,31]? (`*`

means multiplication,`and, or`

are bitwise operations)

### Part IV: Read the Source Code and Write the Output

5 scores for each task.

##### Task 1

```
#include <cstdio>
int main() {
int x;
scanf("%d", &x);
int res = 0;
for (int i = 0; i < x; ++i) {
if (i * i % x == 1) {
++res;
}
}
printf("%d", res);
return 0;
}
```

The input is: `15`

What's the output?

##### Task 2

```
#include <cstdio>
int n, d[100];
bool v[100];
int main() {
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
scanf("%d", d + i);
v[i] = false;
}
int cnt = 0;
for (int i = 0; i < n; ++i) {
if (!v[i]) {
for (int j = i; !v[j]; j = d[j]) {
v[j] = true;
}
++cnt;
}
}
printf("%d\n", cnt);
return 0;
}
```

The input is: `10 7 1 4 3 2 5 9 8 0 6`

What's the output?

##### Task 3

```
#include <iostream>
using namespace std;
string s;
long long magic(int l, int r) {
long long ans = 0;
for (int i = l; i <= r; ++i) {
ans = ans * 4 + s[i] - 'a' + 1;
}
return ans;
}
int main() {
cin >> s;
int len = s.length();
int ans = 0;
for (int l1 = 0; l1 < len; ++l1) {
for (int r1 = l1; r1 < len; ++r1) {
bool bo = true;
for (int l2 = 0; l2 < len; ++l2) {
for (int r2 = l2; r2 < len; ++r2) {
if (magic(l1, r1) == magic(l2, r2) && (l1 != l2 || r1 != r2)) {
bo = false;
}
}
}
if (bo) {
ans += 1;
}
}
}
cout << ans << endl;
return 0;
}
```

The input is: `abacaba`

What's the output?

##### Task 4

```
#include <cstdio>
using namespace std;
const int N = 110;
bool isUse[N];
int n, t;
int a[N], b[N];
bool isSmall() {
for (int i = 1; i <= n; ++i)
if (a[i] != b[i])
return a[i] < b[i];
return false;
}
bool getPermutation(int pos) {
if (pos > n) {
return isSmall();
}
for (int i = 1; i <= n; ++i) {
if (!isUse[i]) {
b[pos] = i;
isUse[i] = true;
if (getPermutation(pos + 1)) {
return true;
}
isUse[i] = false;
}
}
return false;
}
void getNext() {
for (int i = 1; i <= n; ++i) {
isUse[i] = false;
}
getPermutation(1);
for (int i = 1; i <= n; ++i) {
a[i] = b[i];
}
}
int main() {
scanf("%d%d", &n, &t);
for (int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
for (int i = 1; i <= t; ++i) {
getNext();
}
for (int i = 1; i <= n; ++i) {
printf("%d", a[i]);
if (i == n)
putchar('\n');
else
putchar(' ');
}
return 0;
}
```

The first input is: `6 10 1 6 4 5 3 2`

The second input is: `6 200 1 5 3 4 2 6`

What are the outputs?

### Part V: Program Blank Fill-in

In this part, you are given several programs with their descriptions, but some expressions in the programs are deleted. Fill in the blanks to make the programs work as described.

##### Task 1

You have a 1-indexed permutation *P* of length *n*. We define array *Q* as: *Q*_{i} is the smallest *j* satisfying

Unable to parse markup [type=CF_TEX]

. If such*j*doesn't exist,

*Q*

_{i}=

*n*+ 1 .

For example, if *n* = 5, *P* = {1, 5, 4, 2, 3}, then *Q* = {2, 6, 6, 5, 6}.

The following program reads *P* and calculates *Q* with double-linked list. It is guaranteed that *n* ≤ 10^{5}.

```
#include <iostream>
using namespace std;
const int N = 100010;
int n;
int L[N], R[N], a[N];
int main() {
cin >> n;
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
/*fill in(1)*/;
}
for (int i = 1; i <= n; ++i) {
R[i] = /*fill in(2)*/;
L[i] = i - 1;
}
for (int i = 1; i <= n; ++i) {
L[/*fill in(3)*/] = L[a[i]];
R[L[a[i]]] = R[/*fill in(4)*/];
}
for (int i = 1; i <= n; ++i) {
cout << /*fill in(5)*/ << " ";
}
cout << endl;
return 0;
}
```

##### Task 2

Piggy wants to buy *n* objects. *n* ≤ 1000. He can choose from two shops, A and B.

Each of these *n* objects are sold in both shops. If Piggy buy the i-th object at A shop, the cost is a[i]. If Piggy buy the i-th object at B shop, the cost is b[i]. All the prices are no less than 0 and no more than 10000. If the total price in shop A is no less than 50000, everything Piggy buy at A shop will have a 5% discount. That is, if `total_a >= 50000`

, `result = total_a * 0.95 + total_b`

.

Calculate the minimum amount of money does Piggy need to buy the *n* objects.

```
#include <cstdio>
#include <algorithm>
using namespace std;
const int Inf = 1000000000;
const int threshold = 50000;
const int maxn = 1000;
int n, a[maxn], b[maxn];
bool put_a[maxn];
int total_a, total_b;
double ans;
int f[threshold];
int main() {
scanf("%d", &n);
total_a = total_b = 0;
for (int i = 0; i < n; ++i) {
scanf("%d%d", a + i, b + i);
if (a[i] <= b[i])
total_a += a[i];
else
total_b += b[i];
}
ans = total_a + total_b;
total_a = total_b = 0;
for (int i = 0; i < n; ++i) {
if (/*fill in(1)*/) {
put_a[i] = true;
total_a += a[i];
} else {
put_a[i] = false;
total_b += b[i];
}
}
if (/*fill in(2)*/) {
printf("%.2f", total_a * 0.95 + total_b);
return 0;
}
f[0] = 0;
for (int i = 1; i < threshold; ++i)
f[i] = Inf;
int total_b_prefix = 0;
for (int i = 0; i < n; ++i)
if (!put_a[i]) {
total_b_prefix += b[i];
for (int j = threshold - 1; j >= 0; --j) {
if (/*fill in(3)*/ >= threshold && f[j] != Inf)
ans = min(ans, (total_a + j + a[i]) * 0.95 + /*fill in(4)*/);
f[j] = min(f[j] + b[i], j >= a[i] ? /*fill in(5)*/ : Inf);
}
}
printf("%.2f", ans);
return 0;
}
```

The answer will be translated and published maybe tomorrow.