ACM部分习题答案:
A +
B Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 100972 Accepted Submission(s): 33404
Problem Description
Calculate A + B.
Input
Each line will contain two integers A and B. Process to end of file.
Output
For each case, output A + B in one line.
Sample Input
1 1
Sample Output
2
# include
Int main()
{int x,y,s;
while(scanf("%d %d",&x,&y)!=EOF)
{s=x+y;
printf("%d\n",s);}
return 0;
}
Sum Problem
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 85964 Accepted Submission(s): 19422
Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
Input
The input will consist of a series of integers n, one integer per line.
Output
For each case, output SUM(n) in one line, followed by a blank line. You may assume the result will be in the range of 32-bit signed integer.
Sample Input
1
100
Sample Output
1
5050
# include
int main()
{int n;
long int s;
while(scanf("%d",&n)!=EOF)
{ s=0;
while(n>0)
{s=s+n;
n--;
}
printf("%ld\n\n",s);
}
return 0;
}
A +
B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 58216 Accepted Submission(s): 10500
Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 +
2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
#include
#include
int main()
{ char x[1001],y[1001],z[1001];
int n,i,j,k,m,o;
scanf("%d",&n);
o=n;
while(n--)
{ scanf("%s%s",x,y);
i=strlen(x); j=strlen(y);
for(k=0,m=0;i>0&&j>0;i--,j--)
{
m+=x[i-1]-'0'+y[j-1]-'0';
z[k++]=m%10+'0'; m/=10;
}
for(;i>0;i--)
{
m+=x[i-1]-'0';
z[k++]=m%10+'0';
m/=10;
}
for(;j>0;j--)
{ m+=y[j-1]-'0';
z[k++]=m%10+'0'; m/=10;
}
if(m>0) z[k++]=m%10+'0';
printf("Case %d:\n%s + %s = ",o-n,x,y);
for(;k>0;k--) printf("%c",z[k-1]);
printf("\n"); if(n) printf("\n"); }
return 0;
}
Let the Balloon Rise
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 24082 Accepted Submission(s): 7343
Problem Description
Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.
This year, they decide to leave this lovely job to you.
Input
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) -- the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.
A test case with N = 0 terminates the input and this test case is not to be processed.
Output
For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.
Sample Input
5
green
red
blue
red
red
3
pink
orange
pink
Sample Output
red
pink
#include
#include
void main()
{
int n,i,j,k,t,a[1001];
char h[1001][16];
while(scanf("%d",&n)!=EOF)
{
if(n==0) break;
for(i=0;i<=n;++i) a[i]=0;
t=i=k=0;
while(i for(i=0;i for(j=i+1;j if(strcmp(h[i],h[j])==0) { ++a[i]; if(a[i]>k) { k=a[i]; t=i; } } printf("%s\n",h[t]); } } Elevator Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 12849 Accepted Submission(s): 6646 Problem Description The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop. For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled. Input There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed. Output Print the total time on a single line for each test case. Sample Input 1 2 3 2 3 1 Sample Output 17 41 #include int main(){ int a,b,n,t; while(scanf("%d",&n)!=EOF&&n){ t=0;b=0; t+=n*5;//停的总时间累加起来 while(n--){ scanf("%d",&a); if(a>b)t+=6*(a-b);//与前次比若上升 else if(a b=a; } printf("%d\n",t); } return 0; } Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16322 Accepted Submission(s): 4610 Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit. For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39. Input The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero. Output For each integer in the input, output its digital root on a separate line of the output. Sample Input 24 39 Sample Output 6 3 #include #include int main() { int l,s,i; char a[1000]; while(scanf("%s",a)) { l=strlen(a); s=0; if(l==1&&(strcmp(a,"0")==0) ) break; else {