HDU 5651 xiaoxin juju needs help 逆元
题目链接:
hdu:http://acm.hdu.edu.cn/showproblem.php?pid=5651
bc:http://bestcoder.hdu.edu.cn/contests/contest_showproblem.php?cid=682&pid=1002
xiaoxin juju needs help
As we all known, xiaoxin is a brilliant coder. He knew palindromic strings when he was only a six grade student at elementry school.
This summer he was working at Tencent as an intern. One day his leader came to ask xiaoxin for help. His leader gave him a string and he wanted xiaoxin to generate palindromic strings for him. Once xiaoxin generates a different palindromic string, his leader will give him a watermelon candy. The problem is how many candies xiaoxin's leader needs to buy?
This problem has multi test cases. First line contains a single integer T(T\leq 20)T(T≤20) which represents the number of test cases. For each test case, there is a single line containing a string S(1 \leq length(S) \leq 1,000)S(1≤length(S)≤1,000).
For each test case, print an integer which is the number of watermelon candies xiaoxin's leader needs to buy after mod 1,000,000,0071,000,000,007.
3
aa
aabb
a
1
2
1
题解:
1、可行性:
统计每个字母出去的次数,如果有两种即以上字母出现的次数为奇数,则一定不可能排出回文串。
2、统计:
由于回文串左右两边必须相同,所以我们考虑一边就可以了(如果为奇数则正中间一个不管就和偶数情况是一样的了)。
则可以转化为排列组合问题,等价于求解: (cnt[i]代表某个字母出现的次数(cnt[i]>0) )
( )%(1e9+7)
由于涉及到除法,要求逆元:
例子:
求a/b(mod n) 如果b与n互质,则求满足bx=1(%n)的一个解x,原式可转化为a/b*bx(%n),即a*x(%n);这样就把除法消除了。
代码:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; const int maxn = + ;
const int mod = 1e9 + ;
typedef long long LL; char str[maxn];
int n; int cnt[];
int tmp[], tot; LL b[maxn];
//预处理出阶乘
void pre() {
b[] = b[] = ;
for (int i = ; i < maxn; i++) b[i] = (b[i - ] * i) % mod;
}
//扩展的欧几里得算法求逆元
void gcd(LL a, LL b, LL &d, LL &x, LL &y) {
if (!b) {
d = a;
x = ; y = ;
}
else {
gcd(b, a%b, d, y, x);//y=x',x=y';
//x=y'; y=x'-(a/b)*y';
y -= (a / b)*x;
}
} void init() {
tot = ;
memset(cnt, , sizeof(cnt));
} int main() {
pre();
int tc;
scanf("%d", &tc);
while (tc--) {
init();
scanf("%s", str);
n = strlen(str);
for (int i = ; i < strlen(str); i++) {
cnt[str[i] - 'a']++;
}
int flag = ;
for (int i = ; i < ; i++) {
if (cnt[i] % ) flag++;
//统计一半的字母出现的次数
if (cnt[i] > ) {
tmp[tot++] = cnt[i] / ;
}
}
if (flag > ) { printf("0\n"); continue; }
LL sum = ;
for (int i = ; i < tot; i++) {
int t = tmp[i];
sum = (sum*b[t]) % mod;
}
LL d, x, y;
gcd(sum, mod, d, x, y);
//x有可能是负数,需要处理成正的。
x = (x%mod + mod) % mod;
LL ans = (b[n / ] * x) % mod;
printf("%lld\n", ans);
}
return ;
}
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