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	`1`	`+Biscuits are only produced at integer points so we need not worry about + 0.5`
	`2`	`+There are (T/A) different time intervals.`
	`3`	`+Each interval contributes B`
	`4`	`+`
	`5`	`+Total biscuits = (T/A)B`
	`6`	`+`
	`7`	`+The order of operations should be correct`
	`8`	`+`
	`9`	`+-----`
	`10`	`+`
	`11`	`+int main()`
	`12`	`+{`
	`13`	`+ int A, B, T;`
	`14`	`+ cin >> A >> B >> T;`
	`15`	`+`
	`16`	`+ int total = (T/A)*B;`
	`17`	`+ cout << total << "\n";`
	`18`	`+`
	`19`	`+ return 0;`
	`20`	`+}`

Lines changed: 39 additions & 0 deletions

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	`1`	`+The effective impact of value is (value[i] - cost[i]).`
	`2`	`+We can be greedy and take all items with a positive effective value`
	`3`	`+`
	`4`	`+-----`
	`5`	`+`
	`6`	`+int main()`
	`7`	`+{`
	`8`	`+ int no_of_elements;`
	`9`	`+ cin >> no_of_elements;`
	`10`	`+`
	`11`	`+ vector <int> value(no_of_elements + 1);`
	`12`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`13`	`+ {`
	`14`	`+ cin >> value[i];`
	`15`	`+ }`
	`16`	`+`
	`17`	`+ vector <int> cost(no_of_elements + 1);`
	`18`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`19`	`+ {`
	`20`	`+ cin >> cost[i];`
	`21`	`+ }`
	`22`	`+`
	`23`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`24`	`+ {`
	`25`	`+ value[i] -= cost[i];`
	`26`	`+ }`
	`27`	`+`
	`28`	`+ int answer = 0;`
	`29`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`30`	`+ {`
	`31`	`+ if(value[i] > 0)`
	`32`	`+ {`
	`33`	`+ answer += value[i];`
	`34`	`+ }`
	`35`	`+ }`
	`36`	`+`
	`37`	`+ cout << answer << "\n";`
	`38`	`+ return 0;`
	`39`	`+}`

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