本文章由 WyOJ Shojo 从洛谷专栏拉取，原发布时间为 2022-10-24 08:14:30

这是一道模拟题。

先建立 $3$ 个数组：

int c[1210][3];\/\/存储第i每个号码三种的个数
char st[210], name[1210][25];\/\/名字

然后判断这个电话号码应该放在哪里

int type() {
	int i, lst = 0;
	bool f1, f2;
	f1 = f2 = true;
	for(i = 1; i < 8; i++) {
		if(i % 3 == 2)
			continue;
		if(st[i] != st[lst])\/\/号码不相等
			f1 = false;
		if(st[i] >= st[lst])\/\/号码不递减
			f2 = false;
		lst = i;
	}
	return f1 + 2 * f2;
}

然后处理读入：

m0 = m1 = m2 = 0;
for(i = 1; i <= n; ++i) {
    scanf("%d%s", &nm, name[i]);
    while(nm--) {
        scanf("%s", st);\/\/读入号码
        c[i][type()]++;\/\/把号码对应的类型+1
    }
}

然后对于 $c_{i,0},c_{i,1},c_{i,2}$ 取个最大值，然后对于每个数，若他等于对应的最大值，就输出他的名字。

完整代码：

#include <bits\/stdc++.h>
using namespace std;
int c[1145][3];
char st[114], name[1145][25];
int type() {
	int i, lst = 0;
	bool f1, f2;
	f1 = f2 = true;
	for(i = 1; i < 8; i++) {
		if(i % 3 == 2)
			continue;
		if(st[i] != st[lst])
			f1 = false;
		if(st[i] >= st[lst])
			f2 = false;
		lst = i;
	}
	return f1 + 2 * f2;
}
int main() {
	int n, nm, i, m0, m1, m2;
	bool first;
	scanf("%d", &n);
	m0 = m1 = m2 = 0;
	for(i = 1; i <= n; ++i) {
		scanf("%d%s", &nm, name[i]);
		while(nm--) {
			scanf("%s", st);
			c[i][type()]++;
		}
	}
	for(i = 1; i <= n; ++i)
		m0 = max(m0, c[i][0]);
	for(i = 1; i <= n; ++i)
		m1 = max(m1, c[i][1]);
	for(i = 1; i <= n; ++i)
		m2 = max(m2, c[i][2]);
	first = true;
	printf("If you want to call a taxi, you should call: ");
	for(i = 1; i <= n; ++i)
		if(c[i][1] == m1) {
			if(first)
				first = false;
			else
				printf(", ");
			printf("%s", name[i]);
		}
	printf(".\n");
	first = true;
	printf("If you want to order a pizza, you should call: ");
	for(i = 1; i <= n; ++i)
		if(c[i][2] == m2) {
			if(first)
				first = false;
			else
				printf(", ");
			printf("%s", name[i]);
		}
	printf(".\n");
	first = true;
	printf("If you want to go to a cafe with a wonderful girl, you should call: ");
	for(i = 1; i <= n; ++i)
		if(c[i][0] == m0) {
			if(first)
				first = false;
			else
				printf(", ");
			printf("%s", name[i]);
		}
	printf(".\n");
	return 0;
}