用字典来解决

#include <iostream>
#include <cstring>
#include <map>
using namespace std;
int main()
{
    string s;
    getline(cin, s);

    map<string, string> dict;
    dict["shi xian"] = "li bai";
    dict["shi sheng"] = "du fu";
    dict["shi kuang"] = "he zhi zhang";
    dict["shi mo"] = "bai ju yi";
    dict["shi fo"] = "wang wei";
    dict["shi gui"] = "li he";
    dict["shi shen"] = "su shi";
    dict["shi hao"] = "liu yu xi";

    cout << dict[s] << endl;
    return 0;
}

分享一个有意思的解法。因为雅号前面全是"shi"，所以直接忽略掉，匹配后面就行了。

#include <stdio.h>
#include <string.h>

int main() {
	char nickname[10];
	scanf("%*s %s", nickname);
	
	if (strncmp(nickname, "xian", 4) == 0) {
		printf("li bai");
	} else if (strncmp(nickname, "sheng", 5) == 0) {
		printf("du fu");
	} else if (strncmp(nickname, "kuang", 5) == 0) {
		printf("he zhi zhang");
	} else if (strncmp(nickname, "mo", 2) == 0) {
		printf("bai ju yi");
	} else if (strncmp(nickname, "fo", 2) == 0) {
		printf("wang wei");
	} else if (strncmp(nickname, "gui", 3) == 0) {
		printf("li he");
	} else if (strncmp(nickname, "shen", 4) == 0) {
		printf("su shi");
	} else if (strncmp(nickname, "hao", 3) == 0) {
		printf("liu yu xi");
	}
	
	return 0;
}

可以用二维数组定义a和b两个，分别对应雅号和名字。

#include<stdio.h>
#include<string.h>
#include<stdlib.h> 
#include<math.h>


int main(){
	char a[8][20]={"shi xian","shi sheng","shi kuang","shi mo","shi fo","shi gui","shi shen","shi hao"};
	char b[8][20]={"li bai","du fu","he zhi zhang","bai ju yi","wang wei","li he","su shi","liu yu xi"};
	char s[50];
	fgets(s,50,stdin);
	s[strcspn(s,"\n")]=0;
	int i;
	for(i=0;i<8;i++)
	if(!strcmp(s,a[i])){printf("%s",b[i]);return 0;}
    return 0;
}