【瞎搞】 ZOJ 1546 Fillword
给出一个N*M的字符矩阵 再给出P行字符
题目上说P行中必定存在与矩阵中。。直接统计矩阵中字符个数
然后减去P行的字符,然后输出剩下的字符(按照字典树)
#include <cstdio> #include <cstdlib> #include <cstring> #include <climits> #include <cctype> #include <cmath> #include <string> #include <sstream> #include <iostream> #include <algorithm> #include <iomanip> using namespace std; #include <queue> #include <stack> #include <vector> #include <deque> #include <set> #include <map> typedef long long LL; #pragma comment(linker, "/STACK:1024000000,1024000000") #define pi acos(-1.0) #define lson l, m, rt<<1 #define rson m+1, r, rt<<1|1 typedef pair<int, int> PI; typedef pair<int, PI> PP; #ifdef _WIN32 #define LLD "%I64d" #else #define LLD "%lld" #endif //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;} //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;} //inline void print(LL x){printf(LLD, x);puts("");} //inline void read(int &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}} int orz[30]; char s[441]; int main() { int n,m,t; #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); freopen("out.txt", "w", stdout); #endif while(scanf("%d%d%d",&n,&m,&t)!=EOF) { memset(orz,0,sizeof(orz)); for(int i=0;i<n;i++) { scanf("%s",s); for(int j=0;j<m;j++) orz[s[j]-'A']++; } for(int i=0;i<t;i++) { scanf("%s",s); int len=strlen(s); for(int j=0;j<len;j++) orz[s[j]-'A']--; } for(int i=0;i<26;i++) { if(orz[i]>0) while(orz[i]--) printf("%c",'A'+i); } printf("\n"); } return 0; }