常用代码整理:
#include<iostream> #include<cstdio> #include<cstring> #include<string> using namespace std; bool startWith(const string &str,const char* prex) { int pos = str.find(prex); return pos == 0; } bool endWith(const string & str,char* prex) { int pos = str.rfind(prex); return pos == str.length()-strlen(prex); } int main() { char cstr[100] = "abcdefg"; string str(cstr); //字节数 //length与size内部实现与值都是一样的,length只是沿用 //之前的c语言风格 printf("%d %d\n",str.length(),str.size()); //追加 string str2("12345"); str += str2; //方法1 printf("%s\n",str.c_str()); char cstr2[100] = "12345"; str.append(cstr2); //方法2 printf("%s\n",str.c_str()); //子串,substr参数(int pos,int len); str = str.substr(0,strlen(cstr)); printf("%s\n",str.c_str()); //比较,返回值为int,为0相等 int ret = str.compare(str2); printf("%d\n",ret); //查找 int pos = str.find("bcd",0); printf("pos = %d\n",pos); //替代,替代的长度与字符串的长度可以不相等 //对本身的修改 str.replace(0,3,str2); printf("%s\n",str.c_str()); //插入 //对本身的修改 str.insert(2,str); printf("%s\n",str.c_str()); //请后缀判断 cout<<"test\n"; string str5("123abcdg321"); cout<<startWith(str5,"123")<<" "<<startWith(str5,"122")<<endl; cout<<endWith(str5,"321")<<" "<<endWith(str5,"322")<<endl; return 0; }
例题有浙大PAT甲级1058
链接如下:
https://www.patest.cn/contests/pat-a-practise/1058
#include<iostream> #include<cstring> #include<cstdio> #include<string.h> #include<cmath> #include<algorithm> using namespace std; struct ST { int a; int b; int c; }; ST add(ST st1,ST st2) { ST res; int add = 0; res.c = (st1.c+st2.c)%29; add =(st1.c+st2.c)/29; res.b = (st1.b+st2.b+add)%17; add = (st1.b+st2.b+add)/17; res.a = st1.a+st2.a+add; return res; } ST change(string str) { ST st; int index1,index2; index1 = str.find('.'); index2 = str.find('.',index1+1); string s1,s2,s3; s1 = str.substr(0,index1); s2 = str.substr(index1+1,index2-index1-1); s3 = str.substr(index2+1,str.length()-index2-1); //printf("index=%d,%d\n",index1,index2); //printf("s1=%s,s2=%s,s3=%s\n",s1.c_str(),s2.c_str(),s3.c_str()); st.a = atoi(s1.c_str()); st.b = atoi(s2.c_str()); st.c = atoi(s3.c_str()); return st; } int main() { char cs1[20],cs2[20]; scanf("%s%s",cs1,cs2); string str1(cs1),str2(cs2); ST res = add(change(str1),change(str2)); printf("%d.%d.%d\n",res.a,res.b,res.c); return 0; }