# Peking University poj # 1001Exponentiation power,

# Peking University poj # 1001Exponentiation power,

Attached to the original question address: http://poj.org/problem? Id = 1001

First, I analyzed the questions and listed several important points I think:

Next, we will break through them one by one;

• Enter two integers to calculate the product

Because it is a large number, it has exceeded the range that can be calculated by int or long double, so I thought of using the array method for processing.

Save the multiplier to an array, and then use the column-vertical method to calculate and import a single digit. The simulation is as follows:

A [0] a [1] a [2] a [3] a [4]

X B [0] B [1] B [2] B [3] B [4]

----------------------------------------------------------------------------

[0] [4] [1] [4] [2] [4] [3] [4] [4] [4]

[0] [3] [1] [3] [2] [3] [3] [3] [4] [3]

[0] [2] [1] [2] [2] [2] [3] [2] [4] [2]

[0] [1] [1] [1] [2] [1] [3] [1] [4] [1]

[0] [0] [1] [0] [2] [0] [3] [0] [4] [0]

----------------------------------------------------------------------------

C [0] c [1] c [2] c [3] c [4] c [5] c [6] c [7] c [8]

Then, before writing this code, I listed an instance to clarify the purpose of each step:

9 9

X 9 9 // two multiplier arrays are constructed. The array length is the number of digits.

---------------------

81

81

---------------------

81 162 81 // multiplication and Addition before carry

---------------------

81 170 1

---------------------

98 0 1 // carry complete

The following code implements "input two integers, calculate the product:

1 # include <stdio. h> 2 int main () {3 int la, lb, l, I, j; 4 char aa [50], bb [50]; 5 scanf ("% s", aa, bb); 6 la = strlen (aa); 7 lb = strlen (bb); 8 l = la + lb-1; 9 int a [la], B [lb], AB [l], c [l]; 10 for (I = 0; I <la; I ++) {11 a [I] = (int) aa [I]-48; 12} 13 for (I = 0; I <lb; I ++) {14 B [I] = (int) bb [I]-48; 15} // get two multiplier arrays 16 for (I = L-1; I> = 0; I --) {17 AB [I] = 0; 18 for (j = la-1; j> I-lb; j --) {19 if (I-j> = 0 & j> = 0) {20 AB [I] = AB [I] + a [j] * B [I-j]; 21} 22} 23} 24 for (I = 0; I <l; I ++) {25 c [I] = AB [I]; 26} // get 27 printf ("\ n") before carrying AB "); 28 for (I = L-1; I> = 1; I --) {29 c [I] = AB [I] % 10; 30 AB [I-1] = AB [I-1] + (AB [I]-c [I])/10; 31} 32 c [0] = AB [0]; 33 for (I = 0; I <l; I ++) {34 printf ("% d", c [I]); 35} 36 return 0; 37}

• After multiple multiplier values are input, the branch outputs the result.

Here we use a statement: while (scanf ("% s % d", s, & n) = 2) or while (scanf (% s % d ", s, & n )! = EOF)

The function is to jump out of the loop after the input data is complete.

The following code implements "output result of branch after multiple multiplier inputs:

 1 #include <stdio.h> 2 int main() { 3     int x[100],y[100],z[100],i=0,j; 4     while(scanf("%d %d",&x[i],&y[i])==2){ 5         z[i]=x[i]*y[i]; 6         i++; 7     } 8     for(j=0;j<i;j++){ 9         printf("%d\n",z[j]);10     }11     return 0;12 }

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