UVA 10162 - Last Digit
題目連結
題意:求S=(11+22+...NN)%10
思路:打出0-9的每個周期,發現周期為1或2或4、是以S是以20一個周期,打出表後發現20為4,是以對應的40為8,60為2,80為6,100為0,100為1個周期,且為0,是以先把數字mod上100,然後在mod 20求出對應位置。
代碼:
#include <stdio.h>
#include <string.h>
const int Z2[10] = {0, 4, 8, 2 ,6};
int T[10][10], tn[10];
int Z[25];
char str[105];
int solve(int mod) {
int len = strlen(str);
int yu = 0;
for (int i = 0; i < len ;i++) {
yu = (yu * 10 + str[i] - '0') % mod;
}
return yu;
}
int main() {
for (int i = 0; i < 10; i++) {
int tmp = i;
T[i][tn[i]++] = i;
tmp = tmp * i % 10;
while (tmp != i) {
T[i][tn[i]++] = tmp;
tmp = tmp * i % 10;
}
}
for (int i = 1; i <= 20; i++) {
int tmp = i % tn[i % 10] - 1;
if (tmp < 0) tmp += tn[i % 10];
Z[i] = (Z[i - 1] + T[i % 10][tmp]) % 10;
}
/*for (int i = 1; i <= 20; i++)
printf("%d\n", Z[i]);
for (int i = 0; i < 10; i++) {
printf("%d: zhouqi %d:\n", i, tn[i]);
for (int j = 0; j < tn[i]; j++)
printf("%d ", T[i][j]);
printf("\n");
}*/
while (~scanf("%s", str) && str[0] != '0') {
int yu = solve(100);
int ans = (Z[yu % 20] + Z2[yu / 20]) % 10;
printf("%d\n", ans);
}
return 0;
}