C语言使用openssl大数指数求余

1.源码实现

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <openssl/bn.h>

int main()
{
        BN_CTX *r[3];
        BIGNUM *a;
        BIGNUM *b;
        BIGNUM *c;
        BIGNUM *d;
        BIGNUM *e;
        BIGNUM *f;
        BIGNUM *g;
        char u[512+1] = "FE7155A4C441E38D653CC7BB82D9B9A3BFB872CF151B8A19AF4D8E6035BFE6275FFC722ACD29EA9B1FF28B8D134DFAD001738DF7A2F78909DEF606836EE6C6CB";
        char s[512+1] = "FDBAE7C63C5ECA65D27E9C8759AED3B8F6B36F24773FB6A4A03B49026B2D82491732CFDC9F3C77B61B19A945AA2386A7ABD69F23FBBFE18EDB6B7780A3C4CB13";
        char t[512+1] = "C5C671EBC128B4423C6323DF7E43A1DD35EFCFF0F17B09AD1D6B3FCFBBBB685E6C725BCBC3518B9E067B260C2FE5849F24C839C6AA799732D8B5F8EDC9F286CB";
        char *x;

        r[0] = BN_CTX_new();

        a = BN_new();
        b = BN_new();
        c = BN_new();
        d = BN_new();

        BN_hex2bn(&a, s);
        BN_hex2bn(&b, t);
        BN_hex2bn(&c, u);

        BN_mod_exp(d, a, c, b, r[0]);   //d=(a^b)%c

        x = BN_bn2hex(d);

        puts(x);

        BN_free(a);
        BN_free(b);
        BN_free(c);
        BN_free(d);

        BN_CTX_free(r[0]);

        free(x);

        return 0;
}

2.编译源码

$ gcc -I$HOME/local/include -o example example.c -L$HOME/local/lib -lssl

3.运行结果

$ ./example
3D17E54FCF88FF4CAAF8780F5FF6DEAB374C223F8EA0D7F4B00DE6F2A601F8D9E2EE49A654B9F3CB23A2F454F811834D064866967DB903766881E2C6EE2F5ABF