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題目:給你12個硬幣編號A-L,有一個和其他的重量不一樣,現在取某些硬幣放到天平兩端;
給你3中不同的測試結果,為哪個是特殊的硬幣是輕了還是重了(一定有唯一解)。
分析:暴力、枚舉。數據較小直接枚舉24中可能性,判斷條件是否成立即可。
說明:最近做題很少啊╮(╯▽╰)╭。
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int coins[12];
char str_a[3][13],str_b[3][13],str_c[3][13];
int state(char s[])
{
if (!strcmp(s, "even")) return 0;
if (!strcmp(s, "up")) return 1;
if (!strcmp(s, "down")) return -1;
return -2;
}
int compare(char a[], char b[])
{
int value_a = 0,value_b = 0;
for (int i = 0; a[i]; ++ i)
value_a += coins[a[i]-'A'];
for (int i = 0; b[i]; ++ i)
value_b += coins[b[i]-'A'];
return value_a-value_b;
}
int main()
{
int n,s;
while (~scanf("%d",&n))
while (n --) {
for (int i = 0; i < 3; ++ i)
scanf("%s %s %s",str_a[i],str_b[i],str_c[i]);
for (int p = 0; p < 12; ++ p)
for (int q = 0; q <= 1; ++ q) {
for (int i = 0; i < 12; ++ i)
coins[i] = 0;
if (q == 1) {
coins[p] = -1;
}else coins[p] = 1;
int flag = 1;
for (int i = 0; i < 3; ++ i) {
if (compare(str_a[i], str_b[i]) != state(str_c[i])) {
flag = 0;
break;
}
}
if (flag) {
for (int i = 0; i < 12; ++ i) {
if (coins[i] == -1 || coins[i] == 1) {
printf("%c is the counterfeit coin and it is ",'A'+i);
if (coins[i] < 0) {
printf("light.\n");
}else printf("heavy.\n");
break;
}
}
}
}
}
return 0;
}
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原文地址:http://blog.csdn.net/mobius_strip/article/details/45094765