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u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11752 Accepted Submission(s): 5028

Problem Description

A simple mathematical formula for e is

where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333

#include <stdio.h>
int main(){
    double sum;
    int n, i, l;
    printf("n e\n");
    printf("- -----------\n");
    printf("0 1\n");
    sum = 1;
    l = 1;
    for (n=1; n<=9; ++n){
        l *= n;
        sum += 1.0/l;
        if (n <= 2){
            printf("%d %g\n", n, sum);
        }
        else{
            printf("%d %.9lf\n", n, sum);
        }
    }
    return 0;
}

Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15522 Accepted Submission(s): 8168

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
0

Sample Output

17
41

#include <stdio.h>
int main(){
    int n, t, l, r;
    while (scanf("%d", &n) == 1){
        if (n == 0){
            break;
        }
        r = 0;
        l = 0;
        while (n-- != 0){
            scanf("%d", &t);
            if (t > l){
                r += (t-l)*6;
            }
            else{
                r += (l-t)*4;
            }
            r += 5;
            l = t;
        }
        printf("%d\n", r);
    }
    return 0;
}

杭电ACM 1001 Sum Problem

Sum Problem

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 108140 Accepted Submission(s): 24596

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 <stdio.h>
int main(){
    int n, sum;
    while (scanf("%d", &n) == 1){
        sum = 0;
        while (n != 0){
            sum += n--;
        }
        printf("%d\n\n", sum);
    }
    return 0;
}

neuxxm's Blog

neuxxm's Blog - 本为贵公子，平生实爱才。感时思报国，拔剑起蒿莱。

杭电ACM 1012 u Calculate e

u Calculate e

杭电ACM 1008 Elevator

Elevator

杭电ACM 1001 Sum Problem

Sum Problem