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我觉得好假啊。
把if else 位置换了,就过了。还有人预处理前 N 项过的,我矩阵快速幂不服,连1e18数据都没有。
因为有循环节,所以标记,也可以直接找循环节更快。
#include <bits/stdc++.h>
#include <cstring>
#define ll long long
using namespace std;
const ll mod = 998244353;
const ll loop = 499122176;
struct Mat {
int n,m;
ll mem[2][2];
Mat(int n, int m) {
this->n = n, this->m = m;
memset(mem, 0, sizeof mem);
return;
}
Mat operator*(const Mat &a)const {
Mat c(n,a.m);
for (int i = 0; i < c.n; i++) {
for (int j = 0; j < c.m; j++) {
for (int k = 0; k < m; k++)
c.mem[i][j] += mem[i][k] * a.mem[k][j];
c.mem[i][j] %= mod;
}
}
return c;
}
};
Mat solve(Mat a,Mat c,char *s) {
Mat b = c;
// printf("%s\n",s);
for(int i=strlen(s)-1; i>=0; i--) {
int cur = s[i] - '0';
while(cur--) b = b*a;
Mat tmp = a*a;
a = tmp*tmp;
a = a*a*tmp;
}
return b;
}
ll num;
ll cf;
map<ll,ll> q;
char s[5000];
int main() {
ll x0,x1,a,b;
ll tmp,loop;
Mat m1(2,2),m2(2,2);
x0 = 0;
x1 = 1;
a = 3;
b = 2;
m1.mem[0][0] = 0;
m1.mem[0][1] = b;
m1.mem[1][0] = 1;
m1.mem[1][1] = a;
m2.mem[0][0] = x0;
m2.mem[1][0] = 0;
m2.mem[0][1] = x1;
m2.mem[1][1] = 0;
scanf("%lld",&num);
scanf("%lld",&cf);
ll f1,f2,f3,ans=0,last;
for(int i=1; i<=num; i++) {
//ltoa(cf,s,10);
if(q[cf]==0) {
sprintf(s,"%lld",(cf));
m1.mem[0][0] = 0;
m1.mem[0][1] = b;
m1.mem[1][0] = 1;
m1.mem[1][1] = a;
m1 = solve(m1,m2,s);
last = m1.mem[0][0];
q[cf] = last;
ans^=last;
cf = cf^(last*last);
} else if(q[cf]!=0) {
last = q[cf];
cf = cf^(q[last]*q[last]);
ans^=last;
continue;
}
}
printf("%lld",ans);
return 0;
}
/*
1 1 2 2
1 8
*/