(题目链接)
题意
12个硬币中有1个是假的，给出3次称重结果，判断哪个硬币是假币，并且判断假币是比真币中还是比真币轻。
Solution
很久以前写的题了，现在翻了翻发现思路还是不错的。
细节
像这种比较水的与字符串相关的题目用string做一些处理会方便很多，然而这道题好像都差不多。
代码
// poj1013#include#include #include #include #include #include #include #define MOD 1000000007#define inf 2147483640#define LL long long#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);using namespace std;inline LL getint() { LL x=0,f=1;char ch=getchar(); while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();} while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();} return x*f;}int a[12];string s1,s2,s3;int main() { int T; scanf("%d",&T); while (T--) { int cnt=0; for (int i=0;i<12;i++) a[i]=0; for (int o=1;o<=3;o++) { cin>>s1>>s2>>s3; if (s3=="up") { for (int j=0;j<(int)s1.size();j++) a[s1[j]-'A']++; for (int j=0;j<(int)s2.size();j++) a[s2[j]-'A']--; cnt++; } if (s3=="down") { for (int j=0;j<(int)s1.size();j++) a[s1[j]-'A']--; for (int j=0;j<(int)s2.size();j++) a[s2[j]-'A']++; cnt++; } if (s3=="even") for (int j=0;j<(int)s1.size();j++) a[s1[j]-'A']=a[s2[j]-'A']=6; } for (int i=0;i<12;i++) { if (a[i]==cnt) printf("%c is the counterfeit coin and it is heavy.\n",i+'A'); if (a[i]==-cnt) printf("%c is the counterfeit coin and it is light.\n",i+'A'); } } return 0;}