recursion, power sum

Author: prabaprakash

the-power-sum.js

@@ -0,0 +1,76 @@
+'use strict';
+
+const fs = require('fs');
+
+process.stdin.resume();
+process.stdin.setEncoding('utf-8');
+
+let inputString = '';
+let currentLine = 0;
+
+process.stdin.on('data', inputStdin => {
+    inputString += inputStdin;
+});
+
+process.stdin.on('end', _ => {
+    inputString = inputString.replace(/\s*$/, '')
+        .split('\n')
+        .map(str => str.replace(/\s*$/, ''));
+
+    main();
+});
+
+function readLine() {
+    return inputString[currentLine++];
+}
+
+function count(S, m, n, power) {
+    // If n is 0 then there is 1 solution 
+    // (do not include any coin) 
+    if (n === 0)
+        return 1;
+
+    // If n is less than 0 then no 
+    // solution exists 
+    if (n < 0)
+        return 0;
+
+    // If there are no coins and n 
+    // is greater than 0, then no 
+    // solution exist 
+    if (m <= 0 && n >= 1)
+        return 0;
+
+    // count is sum of solutions (i) 
+    // including S[m-1] (ii) excluding S[m-1] 
+    return count(S, m - 1, n, power) + count(S, m - 1, n - Math.pow(S[m - 1], power), power);
+}
+
+// Complete the powerSum function below.
+function powerSum(X, N) {
+    let m = 1;
+    let sums = []
+    for (let i = 1; i < X; i++) {
+        if (Math.pow(i, N) > X) {
+            break;
+        } else {
+            sums.push(i);
+        }
+    }
+    console.log(sums)
+    return count(sums, sums.length, X, N);
+}
+
+function main() {
+    const ws = fs.createWriteStream(process.env.OUTPUT_PATH);
+
+    const X = parseInt(readLine(), 10);
+
+    const N = parseInt(readLine(), 10);
+
+    let result = powerSum(X, N);
+
+    ws.write(result + "\n");
+
+    ws.end();
+}