Commit 8f0483d

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+17

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Lines changed: 6 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -1,11 +1,15 @@`
`1`	`1`	`#include<stdio.h>`
`2`	`2`	`#include<limits.h>`
`3`	`3`
	`4`	`+// A utility function to get the maximum of two integers`
`4`	`5`	`int max(int a, int b)`
`5`	`6`	`{`
`6`	`7`	`return (a > b)? a : b;`
`7`	`8`	`}`
`8`	`9`
	`10`	`+`
	`11`	`+ /* Returns the best obtainable price for a rod of length n and`
	`12`	`+ price[] as prices of different pieces */`
`9`	`13`	`int DPcutRod(int price[], int n)`
`10`	`14`	`{`
`11`	`15`	`int val[n+1];`
`@@ -14,11 +18,12 @@ int DPcutRod(int price[], int n)`
`14`	`18`	`for (i = 1; i<=n; i++)`
`15`	`19`	`{`
`16`	`20`	`int max_val = INT_MIN;`
	`21`	`+ // Recursively cut the rod in different pieces and compare different`
	`22`	`+ // configurations`
`17`	`23`	`for (j = 0; j < i; j++)`
`18`	`24`	`max_val = max(max_val, price[j] + val[i-j-1]);`
`19`	`25`	`val[i] = max_val;`
`20`	`26`	`}`
`21`		`-`
`22`	`27`	`return val[n];`
`23`	`28`	`}`
`24`	`29`

Lines changed: 6 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -1,12 +1,18 @@`
`1`	`1`	`#include <stdio.h>`
`2`	`2`
	`3`	`+// Returns true if there is a subset of a[] with sun equal to given sum`
`3`	`4`	`bool DP(int set[], int n, int sum)`
`4`	`5`	`{`
	`6`	`+ // The value of subset[i][j] will be true if there is a`
	`7`	`+// subset of set[0..j-1] with sum equal to i`
`5`	`8`	`bool subset[n+1][sum+1];`
	`9`	`+ // If sum is 0, then answer is true`
`6`	`10`	`for (int i = 0; i <= n; i++)`
`7`	`11`	`subset[i][0] = true;`
	`12`	`+ // If sum is not 0 and set is empty, then answer is false`
`8`	`13`	`for (int i = 1; i <= sum; i++)`
`9`	`14`	`subset[0][i] = false;`
	`15`	`+ // Fill the subset table in botton up manner`
`10`	`16`	`for (int i = 1; i <= n; i++)`
`11`	`17`	`{`
`12`	`18`	`for (int j = 1; j <= sum; j++)`

Lines changed: 5 additions & 5 deletions

Original file line number	Diff line number	Diff line change
`@@ -6,13 +6,13 @@ int main()`
`6`	`6`	`int t; cin >> t;`
`7`	`7`	`int i2, i3, i5;`
`8`	`8`	`i2 = i3 = i5 = 0;`
`9`		`- int ugly[500];`
	`9`	`+ int ugly[500];// To store ugly numbers`
`10`	`10`	`ugly[0] = 1;`
`11`	`11`	`for (int i = 1; i < 500; ++i) {`
`12`		`- ugly[i] = min(ugly[i2]2, min(ugly[i3]3, ugly[i5]*5));`
`13`		`- ugly[i] != ugly[i2]*2 ? : i2++;`
`14`		`- ugly[i] != ugly[i3]*3? : i3++;`
`15`		`- ugly[i] != ugly[i5]*5 ? : i5++;`
	`12`	`+ ugly[i] = min(ugly[i2]2, min(ugly[i3]3, ugly[i5]*5));//get minimum of possible solutions`
	`13`	`+ ugly[i] != ugly[i2]2 ? : i2++;// next_mulitple_of_2 = ugly[i2]2;`
	`14`	`+ ugly[i] != ugly[i3]3? : i3++;// next_mulitple_of_3 = ugly[i3]3`
	`15`	`+ ugly[i] != ugly[i5]5 ? : i5++;// next_mulitple_of_5 = ugly[i5]5;`
`16`	`16`	`}`
`17`	`17`	`int n;`
`18`	`18`	`while(t--) {`

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