打表的大水题。
/* 4541 */
#include <cstdio>
#include <cstdlib>
#include <cstring> int a3[] = {, , , };
int a4[] = {, , , };
int a5[] = {, , , , , };
int a6[] = {, , , , , };
int a7[] = {, , };
int a8[] = {, , , , , , , , , , , , };
int a9[] = {, , , , , , , , , , , , , , , , , , , , , }; int main() {
int t, m, n;
int n3, n4, n5, n6, n7, n8, n9;
int i; #ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif n3 = sizeof(a3)/sizeof(int);
n4 = sizeof(a4)/sizeof(int);
n5 = sizeof(a5)/sizeof(int);
n6 = sizeof(a6)/sizeof(int);
n7 = sizeof(a7)/sizeof(int);
n8 = sizeof(a8)/sizeof(int);
n9 = sizeof(a9)/sizeof(int); scanf("%d", &t);
for (i=; i<=t; ++i) {
scanf("%d %d", &n, &m);
printf("Case #%d: ", i);
--m;
if (n == ) {
if (m >= n3)
printf("-1\n");
else
printf("%d\n", a3[m]);
}
if (n == ) {
if (m >= n4)
printf("-1\n");
else
printf("%d\n", a4[m]);
}
if (n == ) {
if (m >= n5)
printf("-1\n");
else
printf("%d\n", a5[m]);
}
if (n == ) {
if (m >= n6)
printf("-1\n");
else
printf("%d\n", a6[m]);
}
if (n == ) {
if (m >= n7)
printf("-1\n");
else
printf("%d\n", a7[m]);
}
if (n == ) {
if (m >= n8)
printf("-1\n");
else
printf("%d\n", a8[m]);
}
if (n == ) {
if (m < n9)
printf("%d\n", a9[m]);
else if (m == n9)
printf("10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n");
else if (m == n9+)
printf("20000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n");
else if (m == n9+)
printf("60000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n");
else if (m == n9+)
printf("100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\n");
else
printf("-1\n");
}
} return ;
}