The 14th UESTC Programming Contest Final B

B - Banana Watch

Time Limit: 1000/1000MS (Java/Others)     Memory Limit: 262144/262144KB (Java/Others)

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As a famous technology company, Banana Inc. invents Banana Watch, redefining the watch.

While a normal watch has 12 indexes and two or three moving hands, a Banana Watch has n indexes and a moving hand.

The moving hand is at 0 initially, and in 1st second, it turns 1 index clockwise; in 2nd second, it turns 2 indexes clockwise; ... ;

 in ith second, it turns 
i indexes clockwise. When it moves back to 0 exactly, one minute passes (Yes, Banana Inc. also redefines the minute).

How many seconds in the first minute?

Input

One integer n.

3≤n≤106

Sample input and outputOutput

Print the number of seconds in the first minute.

Sample Input	Sample Output
3	2
5	4

Hint

If n=5, in 1st second, the hand moves to 1; in 2nd second, the hand moves to 3; in 3rd second, the hand moves to 1; in 4th second, 

the hand moves to 0. So the answer for n=5 is 4.

My Solution

用sum[i]表示1~i的和，然后，从1 ~ maxn 查找。第一次出现if((sum[i] %= n) == 0) {printf("%d", i); break;}

然后考虑到数据范围。所以第一发有maxn = 2000000 + 1000，然后就过了。

假设用 1000000 + 8可能过也可能过不了。

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn = 2000000 + 1000;
long long sum[maxn];
int main()
{
    int n;
    scanf("%d", &n);
    sum[0] = 0;
    for(int i = 1; i < maxn; i++){
        sum[i] += sum[i-1]+i;
        if((sum[i] %= n) == 0) {printf("%d", i); break;}
    }
    return 0;
}

Thank you!