高精度乘方

2019-11-06 07:12:43

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#include<stdio.h>#include<string.h>#define LEN 200char str1[ 6 ] = { 0 }; //记录初始浮点数的数组int num1[ 6 ] = { 0 },num2[ LEN ] = { 0 }, num[ LEN ] = { 0 }; //记录转换为数字后的各位数void Multiply(){int i, j;for( i = 0; i <= 5; i++ ){for( j = 0; j <= LEN - 1; j++ )num[ i + j ] = num[ i + j ] + num1[ i ] * num2[ j ]; //分别计算各位的乘积，并加到对应的位置上}for( i = 0; i <= LEN - 1; i++ ){if( num[ i ] >= 10 ){num[ i + 1 ] = num[ i + 1] + num[ i ] / 10;num[ i ] = num[ i ] % 10; //判断各位是否需进位}num2[ i ] = num[ i ];num[ i ] = 0;}}int main(){int n, i, j, k, point, place;while( scanf( "%s%d", str1, &n ) == 2 ){k = 0;for( i = 5; i >= 0; i-- ){if( str1[ i ] != '.' ){num2[ k ] = str1[ i ] - '0';num1[ k++ ] = str1[ i ] - '0';}elsepoint = i;//记录小数点位数}for( i = 1; i <= n - 1; i++ )Multiply(); //乘 n 次i = LEN - 1;while( num2[ i ] == 0 )i--;j = 0;while( num2[ j ] == 0 ) j++;place = n * ( 5 - point ); //很巧妙的一部棋，决定了小数点的位置 if( place >= i + 1 ) //输出{PRintf( "." );for( k = place - 1; k >= j; k-- )printf( "%d", num2[ k ] );printf( "/n" );}else{if( j > place - 1 ){for( k = i; k > place - 1; k-- )printf( "%d", num2[k] );}else{for( k = i; k >= j; k-- ){if( k == place - 1 )printf( "." );printf( "%d", num2[ k ] );}}printf( "/n" );}memset( str1, 0, sizeof( str1 ) );memset( num1, 0, sizeof( num1 ) );memset( num2, 0, sizeof( num2 ) );memset( num, 0, sizeof( num ) );}return 0;}

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