链接：https://ac.nowcoder.com/acm/contest/11255/C
来源：牛客网
题目描述
Let LCS(s1,s2)LCS(s1,s2) denote the length of the longest common subsequence (not necessary continuity) of string s1s1 and string s2s2.
Now give you four integers a,b,c,na,b,c,n, you need to find three lowercase character strings s1,s2,s3s1,s2,s3satisfy that ∣s1∣=∣s2∣=∣s3∣=n∣s1∣=∣s2∣=∣s3∣=n
and LCS(s1,s2)=a,LCS(s2,s3)=b,LCS(s1,s3)=cLCS(s1,s2)=a,LCS(s2,s3)=b,LCS(s1,s3)=c.
输入描述:
The first line has four integers a,b,c,na,b,c,n.
0≤a,b,c≤n0≤a,b,c≤n.
1≤n≤10001≤n≤1000.
输出描述:
If there is no solution, output "NO" (without double quotation marks).
If there exists solutions, you only need to output any one: output three lines, the i-th line has one strings sisi.
示例1
输入
复制
1 2 3 4
输出
复制
aqcc
abpp
abcc
示例2
输入
复制
1 2 3 3
输出
复制
NO
简单构造题。不妨设(aleq b leq c)。考虑如下构造：s1为(a)个'a' + (n - a)个'x'，s2为(a)个'a' + (n - a)个'y'，s3的前b个字符和s2相同，然后(c - a)个字符和s1相同，剩下的部分填'z'。如果(n - b < c - a)无解。
最后输出的时候需要根据初始的abc的大小关系确定哪个是真正的s1s2s3。
#include <bits/stdc++.h>
using namespace std;
int n, a, b, c, len[4];
int main() {
cin >> a >> b >> c >> n;
len[1] = a, len[2] = b, len[3] = c;
sort(len + 1, len + 3 + 1);
string s1 = "", s2 = "", s3 = "";
for(int i = 1; i <= n; i++) {
if(i <= len[1]) s1 += 'a';
else s1 += 'x';
}
for(int i = 1; i <= n; i++) {
if(i <= len[1]) s2 += 'a';
else s2 += 'y';
}
bool ok = 1;
for(int i = 1; i <= n; i++) {
if(i <= len[2]) {
s3 += s2[i - 1];
}
else {
if(n - len[2] >= (len[3] - len[1])) {
for(int j = i + 1; j <= i + (n - len[2] - (len[3] - len[1])); j++) {
s3 += 'z';
}
for(int j = 1; j <= len[3] - len[1]; j++) {
s3 += 'x';
}
} else {
ok = 0;
break;
}
break;
}
}
if(!ok) {
cout << "NO";
} else {
if(a <= b && b <= c) {
cout << s1 << endl;
cout << s2 << endl;
cout << s3 << endl;
} else if(a <= c && c <= b) {
cout << s2 << endl;
cout << s1 << endl;
cout << s3 << endl;
} else if(b <= a && a <= c) {
cout << s3 << endl;
cout << s2 << endl;
cout << s1 << endl;
} else if(b <= c && c <= a) {
cout << s3 << endl;
cout << s1 << endl;
cout << s2 << endl;
} else if(c <= a && a <= b) {
cout << s2 << endl;
cout << s3 << endl;
cout << s1 << endl;
} else {
cout << s1 << endl;
cout << s3 << endl;
cout << s2 << endl;
}
}
return 0;
}