本题目要求读入任意字符串，输出字符串的最长数字子串。若无数字字符，则输出“No digits”。 输入格式:
输入任意字符串。 输出格式:
输出字符串的最长数字子串。若无数字字符，则输出“No digits”。 输入样例:
12345
输出样例:
12345
输入样例:
abc12de33445fg6
输出样例:
33445
输入样例:
abcde
输出样例:
No digits
简单题,状态机泡一下就行了
#define debug #ifdef debug #include <time.h> #include "/home/majiao/mb.h" #endif #include <iostream> #include <algorithm> #include <vector> #include <string.h> #include <map> #include <set> #include <stack> #include <queue> #include <math.h> #define MAXN ((int)1e5+7) #define ll long long #define INF (0x7f7f7f7f) #define fori(lef, rig) for(int i=lef; i<=rig; i++) #define forj(lef, rig) for(int j=lef; j<=rig; j++) #define fork(lef, rig) for(int k=lef; k<=rig; k++) #define QAQ (0) using namespace std; #define show(x...) \ do { \ cout << "\033[31;1m " << #x << " -> "; \ err(x); \ } while (0) void err() { cout << "\033[39;0m" << endl; } template<typename T, typename... A> void err(T a, A... x) { cout << a << ' '; err(x...); } namespace FastIO { char print_f[105]; void read() { } void print() { putchar('\n'); } template <typename T, typename... T2> inline void read(T &x, T2 &... oth) { x = 0; char ch = getchar(); ll f = 1; while (!isdigit(ch)) { if (ch == '-') f *= -1; ch = getchar(); } while (isdigit(ch)) { x = x * 10 + ch - 48; ch = getchar(); } x *= f; read(oth...); } template <typename T, typename... T2> inline void print(T x, T2... oth) { ll p3=-1; if(x<0) putchar('-'), x=-x; do{ print_f[++p3] = x%10 + 48; } while(x/=10); while(p3>=0) putchar(print_f[p3--]); putchar(' '); print(oth...); } } // namespace FastIO using FastIO::print; using FastIO::read; int n, m, Q, K; char buf[MAXN]; int main() { #ifdef debug freopen("test", "r", stdin); // freopen("out_main", "w", stdout); clock_t stime = clock(); #endif scanf("%s ", buf); // int lef = -1, rig = -1, ans = 0; #define IS_NUM (buf[i]>='0' && buf[i]<='9') //状态机 //1数字状态 0其他状态 enum now { IN = 1, OUT = 0 }; //在数字串内,和在数字串外 int status = OUT, cnt = 0, ans = 0; string str, tmp; for(int i=0; ; i++) { if(status == IN) { if(IS_NUM) { cnt ++; tmp.push_back(buf[i]); } else { if(tmp.length() > str.length()) str = tmp; ans = max(ans, cnt); tmp.clear(); cnt = 0; } } else { if(IS_NUM) { status = IN; ++ cnt; tmp.push_back(buf[i]); } } if(!buf[i]) { if(tmp.length() > str.length()) str = tmp; ans = max(ans, cnt); break; } } // printf("%d %s\n", ans, str.data()); printf("%s\n", str.length() ? str.data() : "No digits"); #ifdef debug clock_t etime = clock(); printf("rum time: %lf 秒\n",(double) (etime-stime)/CLOCKS_PER_SEC); #endif return 0; }