HDU 2065 "red virus" problem exponential functions

Note:

(1 + x/1! + X ^ 2/2! + ''+ X ^ n/n !) ^ 2*(1 + x ^ 2/2! + ''') ^ 2

From e ^ x = 1 + x/1! + X ^ 2/2! + '''

Original formula = e ^ (2 * X) * (E ^ x + e ^ (-x)/2) ^ 2

= (1/4) * (E ^ (2 * x) + 1) ^ 2

= (1/4) * (E ^ (4 * x) + 2 * E ^ (2 * x) + 1)

= (1/4) * (SIA (4 ^ N) * (x ^ n/n !) + 2 * SIA (2 ^ N) * (x ^ n/n !) + 1)

We can see from the above formula:

X ^ n/n! The coefficient is (4 ^ n + 2*2 ^ n + 1)/4 = 4 ^ (n-1) + 2 ^ (n-1) + 1/4

For this question, you only need to calculate (4 ^ (n-1) + 2 ^ (n-1) % 100.

In this example, we will not mention the design of the remainder for large numbers. We have used the binary value of the index ~~

/*
 * hdu2065.c
 *
 *  Created on: 2011-9-10
 *      Author: bjfuwangzhu
*/

#include<stdio.h>
#define LL long long
#define nmax 100
int modular_exp(int a, LL b) {
int res;
    res = 1;
while (b) {
if (b & 1) {
            res = res * a % nmax;
        }
        a = a * a % nmax;
        b >>= 1;
    }
return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("data.in", "r", stdin);
#endif
int t, i;
    LL n;
while (scanf("%d", &t) != EOF,t) {
for (i = 1; i <= t; i++) {
            scanf("%I64d", &n);
            printf("Case %d: %d\n", i,
                    (modular_exp(4, n - 1) + modular_exp(2, n - 1)) % nmax);
        }
        printf("\n");
    }
return 0;