Question C - Subparagraph Summation
Gives an array of length N, makes Q queries, and queries the sum of all elements in a subsection of length l starting from the first element.
For example, 1 137 9-1, query for subsections and 1 {37 9}-1 with a starting length of 3 for the second element. 3 + 7 + 9 = 19, output 19.
Input
Line 1: A number N, N is the length of the array (2 <= N <= 50000). Lines 2 to N + 1: N elements of the array. Line N + 2 (-10^9 <= N[i] <= 10^9): 1 number Q, Q is the number of queries. Lines N + 3 to N + Q + 2: 2 numbers per line, i, l (1 <= I <= N, I + l <= N)
Output
A total of Q rows, corresponding to the calculation results of Q queries.
Sample Input
5 1 3 7 9 -1 4 1 2 2 2 3 2 1 5
Sample Output
4 10 16 19
#include<stdio.h>
long long sum[1000000],x;
int main()
{
int n;
scanf("%d",&n);
sum[0]=0; //Don't forget to write
for(int i=1;i<=n;i++)
{
scanf("%lld",&x);
sum[i]=sum[i-1]+x;
}
scanf("%d",&n);
int a,b;
for(int i=1;i<=n;i++)
{
scanf("%d%d",&a,&b);
printf("%lld\n",sum[a+b-1]-sum[a-1]);
}
return 0;
}Question D - Sorting by ASCII Code
After three characters are input, the ASCII codes of each character are output in order from smallest to largest.
Input
There are several groups of input data, each on a single line, consisting of three characters with no spaces between them.
Output
For each set of input data, the output line is separated by a space in the middle of the characters.
Sample Input
qwe asd zxc
Sample Output
e q w a d s c x z
#include<stdio.h>
#include<string.h>
int main()
{
char a,b,c,temp;
while(~scanf("%c%c%c",&a,&b,&c))
{
getchar();
if(a>b)
{
temp=a;
a=b;
b=temp;
}
if(a>c)
{
temp=a;
a=c;
c=temp;
}
if(b>c)
{
temp=b;
b=c;
c=temp;
}
printf("%c %c %c\n",a,b,c);
//printf("\n");
}
return 0;
} Question E - Achievement Conversion
Enter a percentage score t and convert it to the corresponding grade with the following rules:
90~100 is A;
80~89 is B;
70-79 is C;
60~69 is D;
0~59 is E;
Input
The input data has multiple groups, each of which takes up one row and consists of an integer.
Output
For each set of input data, output one row. If the input data is not in the range of 0-100, output a line:'Score is error!'.
Sample Input
56 67 100 123
Sample Output
E D A Score is error!
#include<stdio.h>
int main()
{
int x;
while(scanf("%d",&x)!=EOF)
{
if(x>=90&&x<=100)
printf("A\n");
else if(x>=80&&x<=89)
printf("B\n");
else if(x>=70&&x<=79)
printf("C\n");
else if(x>=60&&x<=69)
printf("D\n");
else if(x>=0&&x<=59)
printf("E\n");
else
printf("Score is error!\n");
}
return 0;
}
Question F - What day?
Given a date, the output date is the day of the year.
Input
There are several groups of input data, each group occupies one row, and the data format is YYYY/MM/DD. See sample input specifically. In addition, you can be sure that all input data is legitimate.
Output
For each set of input data, an output line indicates the date is the day of the year.
Sample Input
1985/1/20 2006/3/12
Sample Output
20 71
#include<stdio.h>
int main()
{
int year,month,day;
int leap;
int count;
while(scanf("%d/%d/%d",&year,&month,&day)!=EOF)
{
count=0;
if((year % 400 == 0) || ((year % 100 != 0) && (year % 4 == 0)))//Leap year
{
leap=29;
}
else
{
leap=28;
}
month=month-1;
switch(month)
{
case 11:count+=30;
case 10:count+=31;
case 9:count+=30;
case 8:count+=31;
case 7:count+=31;
case 6:count+=30;
case 5:count+=31;
case 4:count+=30;
case 3:count+=31;
case 2:count+=leap;
case 1:count+=31;
}
count+=day;
printf("%d\n",count);
}
return 0;
}
G Question - Product of Odd Numbers
Give you n integers and find the product of all the odd numbers in them.
Input
The input data contains multiple test instances, each of which takes up one row, and the first number of each row is n, which means there are n in total for this set of data, followed by n integers. You can assume that there must be at least one odd number for each set of data.
Output
Output the product of all odd numbers in each array, and one line for the test case.
Sample Input
3 1 2 3 4 2 3 4 5
Sample Output
3 15
#include<stdio.h>
int main()
{
int n,i,a;
int m=1;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
{
scanf(" %d",&a);
if(a%2==1)
m*=a;
}
printf("%d\n",m);
m=1;
}
} H Question - Decision of Prime Numbers
For the expression n^2+n+41, when n takes an integer value in the (x,y) range (including x,y) (-39<=x<y<=50), determine whether the values of the expression are all prime numbers.
Input
The input data has multiple groups, each of which takes up one row and consists of two integers, X and y. When x=0 and y=0, the input ends and the row is not processed.
Output
For each given range of values, if the value of the expression is a prime number, output "OK", otherwise output "Sorry", with one line for each set of outputs.
Sample Input
0 1 0 0
Sample Output
OK
#include<stdio.h>
int main()
{
int n,i,x,y;
while(scanf("%d%d",&x,&y)!=EOF&&(x||y))//-39<=x<y<=50
{
int isprime=1;
for(n=x;n<=y;n++)
{
for(i=2;i<(n*n+n+41);i++)
{
if((n*n+n+41)%i==0)
{
isprime=0;
break;
}
}
}
if(isprime)
printf("OK");
else
printf("Sorry");
printf("\n");
}
return 0;
}Title I - The Story of Ta Tao
This puzzle, oh ^-^
The situation was as follows:
Monkey eats more than half of the peaches on the first day, eats more than half of the remaining peaches on the next day, and then eats more than half of the remaining peaches on the previous day every day, leaving only one peach when he prepares to eat on the next day. Clever you, please help Wukong figure out how many peaches he had when he started eating on the first day.
Input
The input data has multiple groups, each of which takes up one row, and contains a positive integer n (1 < n < 30), which means that when only one peach is left, it occurs on the n-th day.
Output
For each set of input data, the total number of peaches at the beginning of the first day of eating is output, with one row per test instance.
Sample Input
2 4
Sample Output
4 22
#include<stdio.h>
int main()
{
int n;//Days entered, 1<n<30
int i;//Control loop variable
int x=1;//Total number of peaches
while(~scanf("%d",&n))//Enter days n
{
for(i=1;i<n;i++)
{
x=(x+1)*2;
}
printf("%d\n",x);//Number of output peaches x
x=1;//Keep in mind that once the loop is completed, the total number should be cleared, otherwise the next set of calculations will be incorrect
}
}
//n x
//1 1
//2 4
//3 10
//4 22