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78.cpp
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78.cpp
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#include <iostream>
#include <vector>
int main() {
// store result (modulo 10^6 or 10^9 + 7)
std::vector < unsigned long long > partitions;
// degenerated case, there's one partition for an empty pile
partitions.push_back(1);
//#define ORIGINAL
#ifdef ORIGINAL
const long long modulo = 1000000; // 10^6
#else
const long long modulo = 1000000007; // 10^9 + 7
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
#endif
{
unsigned int limit = 100000; // the solution is < 100000, program ab
#ifndef ORIGINAL
std::cin >> limit;
#endif
// fill cache
for (unsigned int n = partitions.size(); n <= limit; n++) {
// sum according to Euler's formula
long long sum = 0;
// all pentagonal numbers where pentagonal(i) <= n
for (unsigned int i = 0;; i++) // abort inside loop
{
// generate alternating numbers +1,-1,+2,-2,+3,-3,...
int alternate = 1 + (i / 2); // generate the digit 1,1,2,2,3,3,...
if (i % 2 == 1)
alternate = -alternate; // flip the sign for every second number
// pentagonal index, "how far we go back" in partitions[]
unsigned int offset = alternate * (3 * alternate - 1) / 2;
// can't go back that far ? (array index would be negative)
if (n < offset)
break;
// add two terms, subtract two terms, add two terms, subtract two terms, ...
if (i % 4 < 2)
sum += partitions[n - offset]; // i % 4 = { 0, 1 }
else
sum -= partitions[n - offset]; // i % 4 = { 2, 3 }
// only the last digits are relevant
sum %= modulo;
}
// note: sum can be temporarily negative
if (sum < 0)
sum += modulo;
#ifdef ORIGINAL
// "divisible by one million" => sum % 1000000 == 0
// last 6 digits (modulo was 10^6) are zero
if (sum == 0)
break;
#endif
partitions.push_back(sum);
}
// print (cached) result
#ifdef ORIGINAL
std::cout << partitions.size() << std::endl;
#else
std::cout << partitions[limit] << std::endl;
#endif
}
return 0;
}