Sicily 1198 Substring

中大sicily编程平台上机考试资料整理 中大sicily编程平台上机考试资料整理,包含考试可能用到的各种算法,很实用! 立即下载

题目要求输出的组合字符串具有最小字典序,其实就是把题目给出的字符串进行排序。考虑输入只有两个字符串s1和s2的情况,要输出s1和s2组合的最小字典序,只需要比较s1+s2和s2+s1这两个字符串哪个的字典序更小即可。扩展到多个字符串的情况也是一样的。

#include <stdio.h>
#include <string.h>
int main()
{
    int t,n;
    int i,j,k;
    char str[25][250];
    char s1[250];
    char s2[250];
    int m;
    char mi[250];

    scanf( "%d",&t );
    for ( i=0;i<t;i++ ) {
        scanf( "%d",&n );
        for ( j=0;j<n;j++ ) {
            scanf( "%s",&str[j] );
        }
        for ( m=0;m<n;m++ ) {
            for ( j=0;j<n-m-1;j++ ) {
               strcpy(s1,str[j]);
               strcpy(s2,str[j+1]);
               strcat(s1,str[j+1]);
               strcat(s2,str[j]);
               k=strcmp(s1,s2);
                
                if ( k>0 ) {
                    strcpy(mi,str[j]);
                    strcpy(str[j],str[j+1]);
                    strcpy(str[j+1],mi);
                } 
            }
        }
        for ( m=0;m<n;m++ ) {
            printf( "%s",str[m] );
        }
        printf( "\n" );
    }
    return 0;
}
告别盲目调参:手把手教你用RKISP2.x Tuner理解AWB、AE、NR等核心模块 本文详细介绍了如何使用RKISP2.x Tuner工具进行ISP调试,重点解析AWB、AE、NR等核心模块的调参技巧。通过RV1126平台的实际操作演示,帮助工程师快速掌握可视化对比实验方法,告别盲目调参,提升图像信号处理效率。 阅读详情

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Sicily 1198 substring及sort方法

#include #include #include using namespace std; bool compare(string a,string b) {     return a+b } int main() {     int t,n;     cin>>t;     while(t!=0)     {         cin>>n;

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sicily 1198. Substring

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sicily1198. Substring

1198. Substring   Constraints Time Limit: 1 secs, Memory Limit: 32 MB Description   Dr lee cuts a string S into N pieces,s[1],…,s[N].    Now, Dr lee gives you these N sub-strings: s[1],…s[N].

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sicily 1198 substring

sort排序,注意ba,b的情况,结果应该是bab而不是bba,只需要在cmp函数里判断b+ba和ba+b的大小即可! 1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 5 using namespace std; 6 7 string s[10...

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