《算法竞赛入门经典》（竖式问题）

来源：互联网发布：阿里云电视盒子破解编辑：程序博客网时间：2024/06/06 04:57

可以用sprinf将一串字符存在一个数组中，然后用strchr判断是否都满足条件。

#pragma warning(disable:4996)#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>#include<map> #include<set>#include<sstream>#include<functional>#include<cassert>#include<list>#include<iterator>#include<utility>#include <stdexcept>  #include <sstream>#include <fstream> #include<unordered_map>#include<unordered_set>using namespace std;using namespace std::placeholders;int main(){    int count = 0;    char s[20], buf[99];    scanf("%s", s);    for (int abc = 111; abc <= 999; abc++)        for (int de = 11; de <= 99; de++)        {        int x = abc*(de % 10);        int y = abc*(de / 10);        int z = abc*de;        sprintf(buf, "%d%d%d%d%d", abc, de, x, y, z);        int ok = 1;        for (int i = 0; i < strlen(buf); i++)            if (strchr(s, buf[i]) == NULL)ok = 0;        if (ok)        {            printf("<%d>\n", ++count);            printf("%5d\nX%4d\n-----\n%5d\n%4d\n-----\n%5d\n\n", abc, de, x, y, z);        }    }    printf("The number of solutions = %d\n", count);    return 0;}

用string很慢

#pragma warning(disable:4996)#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<algorithm>#include<iostream>#include<time.h>#include<map> #include<set>#include<sstream>#include<functional>#include<cassert>#include<list>#include<iterator>#include<utility>#include <stdexcept>  #include <sstream>#include <fstream> #include<unordered_map>#include<unordered_set>using namespace std;using namespace std::placeholders;int vs_main(){    int count = 0;    string s;    cin >> s;    for (int abc = 111; abc <= 999; abc++)        for (int de = 11; de <= 99; de++)        {        int x = abc*(de % 10);        int y = abc*(de / 10);        int z = abc*de;        string ss=to_string(abc) + to_string(de) + to_string(x) + to_string(y) + to_string(z);        int ok = 1;        for (auto i = ss.begin(); i != ss.end();i++)            if (s.find(*i)==string::npos)ok = 0;        if (ok)        {            printf("<%d>\n", ++count);            printf("..%d\nX..%d\n-----\n.%d\n%d.\n-----\n%d\n\n", abc, de, x, y, z);        }        }        printf("The number of solutions = %d\n", count);        return 0;}

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