ACM (UVa) : 357


Data type:
Long Long integer

Algorithm:

number_of_ways [30001];
int coin [] = { 50, 25, 10, 5, 1 };
ways [0] = 1;

for ( int i = 0; i < 5; i++ ) {
for ( int j = coin [i]; j < 30001; j++)
ways [j] += ways [j – coin [i]];
}

output: number_of_ways [n]
// where ‘n’ is given number

// this process shud be done b4 starting to take inputs
// we saved the value dynamically in an array named number_of_ways

if (ways [n] > 1) then
There are … ways to produce … cents change.

otherwise
There is only 1 way to produce … cents change.

Another algorithm:

U may notice, there is a series …
or a set of 5 numbers giving u the same values

Example :
for n = [0, 1, 2, 3, 4] = number of way : 1
for n = [5, 6, 7, 8, 9] = number of ways : 2
for n = [10, 11, 12, 13, 14] = number of ways : 4
for n = [15, 16, 17, 18, 19] = number of ways : 6
for n = [20, 21, 22, 23, 24] = number of ways : 9
for n = [25, 26, 27, 28, 29] = number of ways : 13
for n = [30, 31, 32, 33, 34] = number of ways : 18
for n = [35, 36, 37, 38, 39] = number of ways : 24
for n = [40, 41, 42, 43, 44] = number of ways : 31
for n = [45, 46, 47, 48, 49] = number of ways : 39
and so on …

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