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1、1000A + B ProblemTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 158161Accepted Submission(s): 50186Problem DescriptionCalculate A + B.InputEach line will contain two integers A and B. Process to end of file.OutputFor each case, output A + B in one
2、 line.Sample Input1 1Sample Output2AuthorHDOJStatistic|Submit|Discuss | Note#includeint main() int a,b; while(scanf(%d %d,&a,&b)!=EOF) printf(%dn,a+b);1002A + B Problem IITime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 84367Accepted Submission(s):
3、15966Problem DescriptionI have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.InputThe 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
4、 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.OutputFor each test case, you should output two lines. The first line is Case #:, # means the number of the test case. The s
5、econd 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 Input21 2112233445566778899 998877665544332211Sample OutputCase 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111
6、111111111110AuthorIgnatius.LStatistic|Submit|Discuss | Note#include #include int main() char str11001, str21001; int t, i, len_str1, len_str2, len_max, num = 1, k; scanf(%d, &t); getchar(); while(t-) int a1001 = 0, b1001 = 0, c1001 = 0; scanf(%s, str1); len_str1 = strlen(str1); for(i = 0; i = len_st
7、r1 - 1; +i) ai = str1len_str1 - 1 - i - 0; scanf(%s,str2); len_str2 = strlen(str2); for(i = 0; i len_str2) len_max = len_str1; else len_max = len_str2; k = 0; for(i = 0; i = 0; -i) printf(%d, ci); printf(n); if(t = 1) printf(n); return 0;1005Number SequenceTime Limit: 2000/1000 MS (Java/Others)Memor
8、y Limit: 65536/32768 K (Java/Others)Total Submission(s): 44346Accepted Submission(s): 9722Problem DescriptionA number sequence is defined as follows:f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2) mod 7.Given A, B, and n, you are to calculate the value of f(n).InputThe input consists of mult
9、iple test cases. Each test case contains 3 integers A, B and n on a single line (1 = A, B = 1000, 1 = n = 100,000,000). Three zeros signal the end of input and this test case is not to be processed.OutputFor each test case, print the value of f(n) on a single line.Sample Input1 1 31 2 100 0 0Sample
10、Output25AuthorCHEN, ShunbaoSourceZJCPC2004 RecommendJGShiningStatistic|Submit|Discuss | Note#includeint f200;int main() int a,b,n,i; while(scanf(%d%d%d,&a,&b,&n)&a&b&n) if(n=3) f1=1;f2=1; for(i=3;i=200;i+) fi=(a*fi-1+b*fi-2)%7; if(fi-1=1&fi=1) break; i-=2; n=n%i; if(n=0) printf(%dn,fi); else printf(
11、%dn,fn); else printf(1n); return 0;1008ElevatorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17764Accepted Submission(s): 9490Problem DescriptionThe 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,
