UVA-537 Artificial Intelligence?-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_33354943/article/details/51920641

本文介绍了解决UVA-537问题的方法，该问题是关于从给定句子中提取电压(U)、电流(I)和功率(P)的相关信息，并使用公式P=U*I来计算缺失的值。通过逐字符解析输入并根据单位前缀调整数值大小，最终输出精确到两位小数的答案。

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UVA-537 Artificial Intelligence?

题目大意：从一句话中抽取出已知条件，应用公式 P = U * I，求出未知的元素。

Sample Input

3
If the voltage is U=200V and the current is I=4.5A, which power is generated?
A light-bulb yields P=100W and the voltage is U=220V. Compute the current, please.
bla bla bla lightning strike I=2A bla bla bla P=2.5MW bla bla voltage?

Sample Output

Problem #1
P=900.00W
Problem #2
I=0.45A
Problem #3
U=1250000.00V

解题思路：对字符逐个判断，遇到 p,u,i 时，判断下一个是否是 = ，如果是，以 lf 读取已知条件的数值，运算后输出结果。

//UVA-537 Artificial Intelligence?
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char c;
double u,i,p;
int count = 1;
char str[100000];
int main() {
    int T;
    scanf("%d",&T);
    getchar();
    while (T--) {
        u = 0, i = 0, p = 0;
        int t = 2;
        while (t--) {
            while ((c = getchar()) != '\n' ) {
                if (c == 'I') {
                    if((c = getchar()) == '=') {
                        scanf("%lf",&i);
                        c = getchar();
                        if(c == 'm')
                            i = i * 0.001;
                        if(c == 'k')
                            i = i * 1000;
                        if(c == 'M')
                            i = i * 1000000;
                        break;
                    }
                }
                if(c == 'P') {
                    if((c = getchar()) != '\n') {
                        scanf("%lf",&p);
                        c = getchar();
                        if(c == 'm')
                            p = p * 0.001;
                        if(c == 'k')
                            p = p * 1000;
                        if(c == 'M')
                            p = p * 1000000;
                        break;
                    }
                }
                if(c == 'U') {
                    if((c = getchar()) != '\n') {
                        scanf("%lf",&u);
                        c = getchar();
                        if(c == 'm')
                            u = u * 0.001;
                        if(c == 'k')
                            u = u * 1000;
                        if(c == 'M')
                            u = u * 1000000;
                        break;
                    }
                }
            }
        }
        gets(str);
        printf("Problem #%d\n",count++);
        if (u == 0)
            printf("U=%.2lfV",p/i);
        if (i == 0)
            printf("I=%.2lfA",p/u);
        if (p == 0)
            printf("P=%.2lfW",u*i);
        printf("\n\n");
    }
    return 0;
}