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1. Problem description
70044 And 113148 What is the greatest common divisor of ?
Answer submission
This is a question to fill in the blanks , You just need to calculate the results and submit them . The result of this question is an integer , Only fill in this integer when submitting the answer , If you fill in extra content, you will not be able to score .
answer
5388
#include<stdio.h> int main() { int a=70044,b=113148; while(a!=b) { if(a>b) a=a-
b; else b=b-a; } printf("%d",a); return 0; }
2. Problem description
No more than 19000 In a positive integer , And 19000 What is the number of Coprime numbers ?
Answer submission
This is a question to fill in the blanks , You just need to calculate the results and submit them . The result of this question is an integer , Only fill in this integer when submitting the answer , If you fill in extra content, you will not be able to score .
answer
7200
#include<stdio.h> int dfs(int a) { if(a%2!=0&&a%5!=0&&a%19!=0) return 1; return
0; } int main() { int n=19000,i,cns=0; for(i=1;i<=19000;i++) { if(dfs(i)==1) {
cns++; } } printf("%d",cns); return 0; }
3. Problem description
A tree 10 Layer binary tree , How many nodes are included at most ?
Note that when a binary tree has only one node, it is a layer .
Answer submission
This is a question to fill in the blanks , You just need to calculate the results and submit them . The result of this question is an integer , Only fill in this integer when submitting the answer , If you fill in extra content, you will not be able to score .
answer
1023
1+2+4+8+16+32+64+128+256+512;
4. Problem description
Hexadecimal number, please 1949 What is the corresponding decimal number ? Note in particular that hexadecimal is given , It's decimal .
Answer submission
This is a question to fill in the blanks , You just need to calculate the results and submit them . The result of this question is an integer , Only fill in this integer when submitting the answer , If you fill in extra content, you will not be able to score .
answer
6473
Sum equals 16*16*16*1+16*16*9+16*4+9;
*
Problem description
Xiao Ming doesn't like numbers very much 2, Include those digits that contain numbers 2 Number of . If the digits of a number do not contain numbers 2, Xiao Ming calls it clean number .
Where is the integer 1 to n in , How many are there ?
Input format
The first line of input contains an integer n.
Output format
The output line contains an integer , Indicates the answer .
sample input
30
sample output
18
Scale and agreement of evaluation cases
about 40% Evaluation use cases for ,1 <= n <= 10000.
about 80% Evaluation use cases for ,1 <= n <= 100000.
For all profiling cases ,1 <= n <= 1000000.
#include<stdio.h> int dfs(int a) { int xb=1,t; while(a>=1) { t=0; t=a%10; if(t
==2) { xb=0; return 0; } a=a/10; } if(xb) return 1; } int main() { long long n,
cns=0,i; scanf("%lld",&n); for(i=1;i<=n;i++) { if(dfs(i)==1) { cns++; } } printf
("%lld",cns); return 0; }
6. Problem description
In sequence a_1, a_2, …, a_n in , Define two elements a_i and a_j The distance is
|i-j|+|a_i-a_j|, That is, the absolute value of the distance of the element subscript plus the difference of the element value , among |x| express x Absolute value of .
Given a sequence , Please find out the maximum element distance between elements .
Input format
The first line of input contains an integer n.
The second line contains n Integer a_1, a_2, …, a_n, Adjacent integers are separated by spaces , Represents a given sequence of numbers .
Output format
The output line contains an integer , Indicates the answer .
sample input
5
9 4 2 4 7
sample output
9
Example description
a_1 and a_3 The distance is |1-3|+|9-2|=9.
Scale and agreement of evaluation cases
about 50% Evaluation use cases for ,2 <= n <= 100,0 <= Number in sequence <= 1000.
For all profiling cases ,2 <= n <= 1000,0 <= Number in sequence <= 10000.
C
#include<stdio.h> #include<math.h> int main() { int n,c; scanf("%d",&n); int a[
n]={0},i,j; for(i=0;i<n;i++) { scanf("%d",&a[i]); } int max=0,sum,h; for(i=0;i<n
;i++) { for(j=i+1;j<n;j++) { h=0;sum=0; h=fabs(i-j); sum=a[i]-a[j]; sum=fabs(sum
+h); max=(sum>max?sum:max); } } printf("%d",max); return 0; }
7. Problem description
In sequence a[1], a[2], …, a[n] in , If a[i] < a[i+1] < a[i+2] < … < a[j], Then called a[i] to
a[j] Is an increasing sequence , Count Reg j-i+1.
Given a sequence , How long is the longest increment sequence in the sequence .
Input format
The first line of input contains an integer n.
The second line contains n Integer a[1], a[2], …, a[n], Adjacent integers are separated by spaces , Represents a given sequence of numbers .
Output format
The output line contains an integer , Indicates the answer .
sample input
7
5 2 4 1 3 7 2
sample output
3
Scale and agreement of evaluation cases
about 50% Evaluation use cases for ,2 <= n <= 100,0 <= Number in sequence <= 1000.
For all profiling cases ,2 <= n <= 1000,0 <= Number in sequence <= 10000.
#include<stdio.h> int main() { int n; int a[101]; int i,j,cns=0,max=0; scanf(
"%d",&n); for(i=0;i<n;i++) { scanf("%d",&a[i]); } for(i=0;i<n;i++) { cns=0; for(
j=i+1;j<n;j++) { if(a[j]>a[j-1]) { cns++; } else break; } cns=cns+1; max=(cns>
max?cns:max); } printf("%d",max); return 0; }
8. Problem description
Given a word , Please count the number of vowels in this word , How many consonants .
Vowels include a, e, i, o, u, Five in total , Others are consonants .
Input format
Enter a line , Contains a word , Words contain only lowercase letters .
Output format
Output two lines , The first line contains an integer , Indicates the number of vowels .
The second line contains an integer , Indicates the number of consonants .
sample input
lanqiao
sample output
4
3
Scale and agreement of evaluation cases
For all profiling cases , The number of letters in a word does not exceed 100.
#include<stdio.h> #include<string.h> int main() { char a[101]; int i,l,cy=0,cf=
0; gets(a); l=strlen(a); for(i=0;i<l;i++) { if(a[i]=='a'||a[i]=='e'||a[i]=='i'||
a[i]=='o'||a[i]=='u') { cy++; } else cf++; } printf("%d\n",cy); printf("%d\n",cf
); return 0; }
9. Problem description
Xiao Ming practices Kung Fu every day , The plum blossom stake is an important item in practice .
The plum blossom stakes of Xiao Ming's practice are arranged in a row n that 's ok m column , The distance between two adjacent rows is 1, The distance between two adjacent columns is also 1.
Xiao Ming stands on the second floor 1 Line number 1 Column , He's going to the third floor n Line number m Column . Xiao Ming has been practicing for some time , He can now move no more than one step d Distance （ Straight line distance ）.
Xiao Ming wants to know , Without dropping the plum blossom pile , How many steps can you move to the target .
Input format
The first line of input contains two integers n, m, Represents the number of rows and columns of plum blossom piles respectively .
The second line contains a real number d（ Contains at most one decimal place ）, Indicates the distance Xiao Ming can move in one step .
Output format
Output an integer , Indicates the minimum number of steps Xiao Ming can take to reach the target .
sample input
3 4
1.5
sample output
3
Scale and agreement of evaluation cases
about 30% Evaluation use cases for ,2 <= n, m <= 20,1 <= d <= 20.
about 60% Evaluation use cases for ,2 <= n, m <= 100,1 <= d <= 100.
For all profiling cases ,2 <= n, m <= 1000,1 <= d <= 100.
#include<stdio.h> #include<math.h> int main() { int n,m,a=1,b=1; float d; scanf
("%d%d",&n,&m); scanf("%f",&d); float xb=1.414214; int cns=0; if(d>xb) { while(a
<=n&&b<=m) { a++;b++; cns++; } } else cns=n-a+m-b; printf("%d",cns); return 0; }
10. Problem description
Xiao Ming built a castle with building blocks .
for convenience , Xiaoming uses a cube of the same size when building , In one n that 's ok m On the grid of columns , Each building block just occupies a small square of the grid .
of course , Xiao Ming's castle is not flat , It's three-dimensional . Xiao Ming can build blocks on other blocks . When the building blocks on a grid are higher , It's a tower , When there are no blocks on a grid , It's a flat ground .
Xiaoming's castle can be represented by the number of building blocks on each grid . for example , Here is a castle .
9 3 3 1
3 3 3 0
0 0 0 0
There are open spaces to the South and east of the castle , There is a big house in the northwest , There is also a high tower in the northwest corner , There is a garage in the northeast corner .
Now? , Gegwu is coming to destroy Xiaoming's castle , He used magic to flood Xiao Ming's castle .
If the height of the water is 1, The building blocks close to the ground will be flooded , In the above example , yes 7 This building block will be flooded .
If the height of the water is 2, More blocks will be flooded , In the above example , yes 13 This building block will be flooded .
Given Xiao Ming's castle map , Excuse me? , The height of water is 1, 2, 3, …, H Time , How many blocks will be flooded .
Input format
The first line of input contains two integers n, m.
next n that 's ok , Each line m Integer , Indicates the number of layers of building blocks at each position in Xiaoming's castle .
Next, include an integer H, Indicates the upper limit of water height .
Output format
output H that 's ok , One integer per line . The first i An integer representing the height of the water is i Number of building blocks flooded during construction .
sample input
3 4
9 3 3 1
3 3 3 0
0 0 0 0
10
sample output
7
13
19
20
21
22
23
24
25
25
Scale and agreement of evaluation cases
about 40% Evaluation use cases for ,1 <= n, m <= 100,1 <= H <= 100, The number of layers of building blocks shall not exceed 100;
about 70% Evaluation use cases for ,1 <= n, m <= 1000,1 <= H <= 1000, The number of layers of building blocks shall not exceed 1000;
For all profiling cases ,1 <= n, m <= 1000,1 <= H <= 100000, The number of layers of building blocks shall not exceed 1000000000.
#include<stdio.h> int main() { int n,m; scanf("%d%d",&n,&m); int a[n][m]; int i
,j; for(i=0;i<n;i++) { for(j=0;j<m;j++) { scanf("%d",&a[i][j]); } } int h,l;
scanf("%d",&h); int cns=0; for(l=1;l<=h;l++) { for(i=0;i<n;i++) { for(j=0;j<m;j
++) { if(a[i][j]>=1) { a[i][j]--; cns++; } } } printf("%d\n",cns); } return 0; }
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