Another Sum Problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1110 Accepted Submission(s): 292
Problem Description
FunnyAC likes mathematics very much. He thinks mathematics is very funny and beautiful.When he solved a math problem he would be very happy just like getting accepted in ACM.Recently, he find a very strange problem.Everyone know that the sum of sequence from 1 to n is n*(n + 1)/2. But now if we create a sequence which consists of the sum of sequence from 1 to n. The new sequence is 1, 1+ 2, 1+2+3, .... 1+2+...+n. Now the problem is that what is the sum of the sequence from1 to 1+2+...+n .Is it very simple? I think you can solve it. Good luck!
Input
The first line contain an integer T .Then T cases followed. Each case contain an integer n (1 <= n <= 10000000).
Output
For each case,output the sum of first n items in the new sequence. Because the sum is very larger, so output sum % 20090524.
Sample Input
3
1
24
56
Sample Output
1
2600
30856
Source
HDU 2009-5 Programming Contest
Recommend
lcy
題意 ：
已知 1 2 3 4 5 6 7 8 9 。。。。n
前n項和 1 3 6 10 15 21
求x 1 4 10 20 35 56
輸入n 輸出x 即求前n項和的前n項和
思路：
假設x=sn
則 sn-sn-1=n(n+1)/2; s1=1；
求sn的通項公式 通過累加法 之後化簡 可求得公式為Sn = n(n+1)(n+2)/6;
對sn求余 我們可以分成2部分 即n(n+1) 和 (n+2)
注意 本題中要保證n(n+1)(n+2)能被6整除 因為sn一定是個整數
所以求余的時候要這樣求余
s1 = (n(n+1))% (20090524*6);
余數裡包含6是保證後面的式子可以被6整除。
s2 = (s1*(n+2)/6)%20090524;
另外本題有個坑爹的地方就是輸入只能用I64d 不能用lld 否則就錯
by hnust_xiehonghao
[cpp]
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long n;
scanf("%I64d",&n);
if(n == 1)
{
printf("1\n");
continue;
}
long long ans = (n*(n+1))%(20090524*6);
ans = (ans*(n+2)/6)%20090524;
printf("%lld\n",ans);
}
return 0;
}
