Calculate the sum of the numbers of 2 ^ 1000/2 ^ 10000

2 ^ 1000 in hexadecimal notation:

2 hexadecimal: 1000 ...... (1000 0 in total)

8 hexadecimal: 2000... (333 0 in total)

Hexadecimal: 10000... (a total of 250 0)

 

The binary to decimal calculation process is equal to: 2*2*2... (a total of 1000 2 multiplied)

Considering that the result of multiplication is relatively large = (2 ^ 10) ^ 100 = (1024) ^ 100> = (1000) ^ 100 = 1000 .. (the last 300 digits are 0)

If accurate calculation is performed, the existing types cannot be met.

 

Therefore, we generally use char * to store related values and perform related calculations:

2 ^ 2 = 2 + 2 = 4

2 ^ 3 = 4 + 4 = 8

2 ^ 4 = 8 + 8 = 16

2 ^ 5 = 16 + 16 = 32

...

2 ^ 1000 is equivalent to 999 addition calculations. We only need to develop a byte addition char * to simulate the 10-digit addition operation.

 

The following computation is a computation program compiled using G ++. The efficiency of computation using PC is as follows:

2 ^ 1000 computing completed 15 ms

2 ^ 10000 computing completed 2.5 s

 

# Include "stdio. H "# include <vector> # include" time. H "# define n_size 10000 bool plusself (STD: vector <char> & vecbytes) {int nplus = 0; For (INT I = 0; I <vecbytes. size (); I ++) {// use * 2 speed = <1 speed> A + A speed. The speed is about 30% faster, probably because of // 1. no need to find variables // 2. for * 2 optimization, use the <1-bit result int result = vecbytes [I] * 2 + nplus; // int result = vecbytes [I] <1 + nplus; // int result = vecbytes [I] + vecbytes [I] + nplus; // The efficiency of the two methods is very close., The second is a little faster, no more than 3% // 1. The second method adds a comparison judgment, but the nplus adopts a direct value assignment // 2. the first method uses remainder and division, because Char data is short and only 8 bytes are used. In this case, it is faster. // we recommend that you, either method works. The first method is short, and the second method is slightly faster. // vecbytes [I] = Result % 10; // nplus = Result/10; if (result <10) {vecbytes [I] = result; nplus = 0 ;}else {vecbytes [I] = Result-10; nplus = 1 ;}}} int main () {STD: vector <char> vecbytes; vecbytes. resize (n_size/3, 0); vecbytes [0] = 1; for (INT I = 0; I <n_size; I ++) {plusself (vecbytes );} for (INT I = 0; I <vecbytes. size (); I ++) {printf ("% C", vecbytes [I] + '0');} printf ("\ n"); int ntotal = 0; for (INT I = 0; I <vecbytes. size (); I ++) {ntotal + = vecbytes [I];} printf ("ntotal: % d clock: % d \ n", ntotal, clock ()); getchar (); Return 0 ;}

 

 

Calculate the sum of the numbers of 2 ^ 1000/2 ^ 10000