问题描述：

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1 £1 + 1 50p + 2 20p + 1 5p + 1 2p + 3 1p

How many different ways can £2 be made using any number of coins?

实现：

思路：

1.先解决1分，2分，5分的情况

2.递归的思路解决其他情况

(function(){

// prepare 

Array.prototype.addRange = function(arr){
if(arr == undefined) { return new Array(); }

var ret  = new Array();
for(var i = 0 ;i < this.length ;i++){
ret.push(this[i]);
}

for(var i = 0 ;i < arr.length; i++){
ret.push(arr[i]);
}

return ret;
}

Array.prototype.constainsArr = function (arr){
var arr1 = new Array();

var isContains = true;
for(var i = 0;i < this.length; i++)arr1.push(this[i]);

var arr2 = new Array();
for(var i = 0;i < arr.length; i++)arr2.push(arr[i]);

var isContain = true;
for(var i = 0;i < arr1.length ; i++){
isContain = true;

arr1[i] = arr1[i].sort(function(a,b){return a-b;});
arr2 = arr2.sort(function(a,b){return a-b;});

for(var j = 0; j< arr1[i].length; j++)
{
if(arr1[i][j]!=arr2[j]) isContain = false;
}

if(isContain) return true;
}

return false;
}


// logic 

var coins = new Array(1, 2, 5, 10, 20, 50, 100 ,200);


var sln = function solve(n){
if(coins.indexOf(n) == -1) throw ex("unexpected coin");

if(n== 1 ) return new Array([1]);
if(n == 2) return new Array([1,1],[2]);
if(n == 5) return new Array([1,1,1,1,1],[1,1,1,2],[1,2,2],[5]);

if(n == 10) {
var arr1 = solve(5);
var combined= new Array();
for(var i = 0 ;i < arr1.length;i++){
for(var j = 0;j < arr1.length;j++){

var ret = arr1[i].addRange(arr1[j]);

if(combined.length == 0 || !combined.constainsArr(ret)){ combined.push(ret); }

}
}

return combined;
}

if(n == 20){
var arr1 = solve(10);
console.log("solved 10 ");
console.log(arr1);

var combined= new Array();
for(var i = 0 ;i < arr1.length;i++){
for(var j = 0;j < arr1.length;j++){

var ret = arr1[i].addRange(arr1[j]);
if(combined.length == 0 || !combined.constainsArr(ret)){ combined.push(ret); }

}
}

return combined;
}

if(n == 50){
var arr1 = solve(20);
var arr2 = solve(10);

var combined= new Array();
for(var i = 0 ;i < arr1.length;i++){
for(var j = 0;j < arr1.length;j++){

var ret = arr1[i].addRange(arr1[j]);
if(combined.length == 0 || !combined.constainsArr(ret)){ combined.push(ret); }


}
}

var combined2 = new Array();
for(var i = 0 ;i < arr2.length;i++){
for(var j = 0;j < combined.length;j++){

var ret = arr2[i].addRange(combined[j]);
if(combined2.length == 0 || !combined2.constainsArr(ret)){ combined2.push(ret); }

}
}
return combined2;

}

if(n== 100){
var arr1 = solve(50);

var combined= new Array();
for(var i = 0 ;i < arr1.length;i++){
for(var j = 0;j < arr1.length;j++){

var ret = arr1[i].addRange(arr1[j]);
if(combined.length == 0 || !combined.constainsArr(ret)){ combined.push(ret); }

}
}

return combined;
}

if(n== 200){
var arr1 = solve(100);

var combined= new Array();
for(var i = 0 ;i < arr1.length;i++){
for(var j = 0;j < arr1.length;j++){

var ret = arr1[i].addRange(arr1[j]);
if(combined.length == 0 || !combined.constainsArr(ret)){ combined.push(ret); }

}
}

return combined;
}

}


var ret = sln(50);
for(var i =0;i < ret.length; i++)
console.log(ret[i]);

})();