The problem of saving money

"Problem description"

Father prepared for the Little Dragon's four college life one-time savings of a sum of money, using the whole deposit 0 to take the way to control the little dragons at the beginning of each month to take 1000 yuan to prepare for this month use. Assuming that the bank deposits 0 per annum at an annual interest of 1.71%, please figure out how much money the father needs to deposit at least.

"Problem Analysis"

This problem is a typical recursive problem, the analysis of the process of saving money and to withdraw funds, we can adopt the method of inverse push. 4 years and 48 months, 1000 yuan a month, the last one months just take it out. We can use an array to hold the remaining amount of money for each month, then the last one months with a 1000, that is, the 48th month in the array value is 1000.

The 47th month Passbook money is: Take away 1000 yuan living expenses + next month 1000 months of principal, namely:

1000+ 48th Month's money/(1+0.00171/12)

The sum of the 46th, three 、......、 1th months can be calculated by analogy:

The money in the passbook for the 46th month is: 1000+ the 47th month (1+0.00171/12)

The money in the passbook for the 45th month is: 1000+ the 46th month (1+0.00171/12)

......

The money in the passbook for the 1th month is: 1000+ the 2nd month (1+0.00171/12)

The amount of the initial deposit can be calculated by the above recursive push.

"Program Code"

1  Public classCQWT2 {  3      Public Static  Final  Doublemoneyrate=0.0171;//Deposit Rate4      Public Static voidMain (string[] args)5     {  6         //define an array of length 48 to fill the remaining deposits at the beginning of each month7         Doublemoney[]=New Double[48]; 8           9         //1000 Yuan at the beginning of the last one monthsTenmoney[47]=1000;  OneSystem.out.printf ("Number of remaining deposits at the beginning of 48:%.2f\n", money[47]);  A         //the remaining deposits in the previous one months are rolled out by cyclic reversal -          for(inti=47;i>0;i--)   -         {   themoney[i-1]=1000+money[i]/(1+MONEYRATE/12);  -System.out.printf ("Number of remaining deposits at the beginning of the month of%d:%.2f\n", i,money[i-1]);  -         }   -         //calculate the amount of money that was originally deposited, that is, the first remaining deposit +System.out.printf ("\ n originally to deposit%.2f yuan. ", money[0]);  -     }   +}

"Run Results"

The problem of saving money