UVa 537 - Artificial Intelligence?

题目：再一句话里面有P，I，U中的2个已知量，求第三个未知量。（P=I*U）

分析：字符串。利用‘=‘定位已知量，然后将‘=‘后面的的数字和单位分别读入处理。

说明：注意单位有m，k，M的前缀，以及小数点的处理；每组输出后面有一个空行。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>

using namespace std;

char Satz[1000];

int main()
{
	int n;
	while ( ~scanf("%d",&n) ) {
		getchar();
		for ( int t = 1 ; t <= n ; ++ t ) {
			gets(Satz);
			double pp = 0.0,ii = 0.0,uu = 0.0;
			int    P = 0,I = 0,U = 0;
			for ( int j,i = 1 ; Satz[i] ; ++ i ) {
				if ( Satz[i] == '=' ) {
					double value = 0.0;
					int    flag = 0;
					for ( j = i+1 ; Satz[j] ; ++ j ) {
						if ( Satz[j] == '.' ) {
							flag = 10;continue;
						}
						if ( Satz[j] < '0' || Satz[j] > '9')
							break;
						if ( flag ) {
							value += (Satz[j]-'0'+0.0)/flag;
							flag *= 10;
						}else {
							value *= 10;
							value += (Satz[j]-'0'+0.0);
						}
					}
					if ( Satz[j] == 'm' ) value *= 0.001;
					if ( Satz[j] == 'k' ) value *= 1000;
					if ( Satz[j] == 'M' ) value *= 1000000;

					if ( Satz[i-1] == 'P' ) {
						pp = value;P = 1;
					}
					if ( Satz[i-1] == 'U' ) {
						uu = value;U = 1;
					}
					if ( Satz[i-1] == 'I' ) {
						ii = value;I = 1;
					}
				}
			}
			printf("Problem #%d\n",t);
			if ( P && U ) printf("I=%.2lfA\n\n",pp/uu);
			if ( P && I ) printf("U=%.2lfV\n\n",pp/ii);
			if ( I && U ) printf("P=%.2lfW\n\n",ii*uu);
		}
	}
	return 0;
}

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