usaco-1-1

Since I don’t have much experience in coding and I just do it for fun, I hope my code can bring you some happiness and understnading here…

This is just USACO 1.1, easy. But the problem afterwards are hard…

Prob 0. 1 + 1 = 2

Just used for testing the input and output form of usaco testing system.

/*
ID:
LANG: C
TASK: test
 */
#include <stdio.h>
#include <stdlib.h>

int main() {
    FILE *fin = fopen("test.in", "r");
    FILE *fout = fopen("test.out", "w");
    int a, b;
    fscanf(fin, "%d %d", &a, &b);
    fprintf(fout, "%d\n", a + b);
    exit(0);
}

Prob 1. Your Ride Is Here

I wrote some not-elegant code for this problem. Anyway, I solved it after all.

• We transfer the letters into numbers.
• Then get the product $mod$ 47.
• If the result is equal to 27, output ‘go’, else output ‘stay’.

/*
ID:
LANG: C
PROG: ride
 */
#include <stdio.h>
#include <stdlib.h>

int ctoi(char bit);

int inscope(char bit);

int transform(char *command);

int main() {
    FILE *fin = fopen("ride.in", "r");
    FILE *fout = fopen("ride.out", "w");
    char command1[6], command2[6];
    fscanf(fin, "%s %s", command1, command2);
    if (transform(command1) == transform(command2))
        fprintf(fout, "GO\n");
    else
        fprintf(fout, "STAY\n");
    exit(0);
}

int ctoi(char bit) { return bit - 'A' + 1; }

int inscope(char bit) { return (bit >= 'A' && bit <= 'Z'); }

int transform(char *command) {
    int target = 1;
    for (int i = 0; i < 6; i++) {
        if (inscope(command[i]))
            target *= ctoi(command[i]);
        else
            break;
    }
    return target % 47;
}

Prob 2. Greedy Gift Givers

A little complicated, but we can still see it through easily.

• We need to establish a relationship between name array and money amount array, so I create a function called find_name.
• Since all the money should be given in an integer form. Then I created a function called give_over and left_over.
  • left_over returns how much you left after you give off money.
  • give_over returns how much you need to give off to others.
• So use find_name to locate the giver and the given, then use give_over and left_over to determine the amount of money change.

/*
ID:
PROG: gift1
LANG: C
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int find_name(char dictionary[][20], int name_quantity, char name_target[]);

int left_over(int quantity, int target_num);

int give_over(int quantity, int target_num);

int main() {
    FILE *fin = fopen("gift1.in", "r");
    FILE *fout = fopen("gift1.out", "w");
    int person_num;
    fscanf(fin, "%d", &person_num);
    char person_name[10][20];
    for (int i = 0; i < person_num; i++)
        fscanf(fin, "%s", person_name[i]);
    int money_pocket[10];
    for (int i = 0; i < person_num; i++)
        money_pocket[i] = 0;
    for (int i = 0; i < person_num; i++) {
        char giver_temp[20];
        fscanf(fin, "%s", giver_temp);
        int giver_plc_temp = find_name(person_name, person_num, giver_temp);
        int quantity, target_num;
        fscanf(fin, "%d %d", &quantity, &target_num);
        money_pocket[giver_plc_temp]
            = money_pocket[giver_plc_temp] - quantity + left_over(quantity, target_num);
        for (int i = 0; i < target_num; i++) {
            char taker_temp[20];
            fscanf(fin, "%s", taker_temp);
            int taker_plc_temp = find_name(person_name, person_num, taker_temp);
            money_pocket[taker_plc_temp] += give_over(quantity, target_num);
        }
    }
    for (int i = 0; i < person_num; i++)
        fprintf(fout, "%s %d\n", person_name[i], money_pocket[i]);
    exit(0);
}

int find_name(char dictionary[][20], int name_quantity, char name_target[]) {
    for (int i = 0; i < name_quantity; i++)
        if (strcmp(dictionary[i], name_target) == 0)
            return i;
    return -1;
}

int left_over(int quantity, int target_num) {
    if (target_num != 0)
        return quantity % target_num;
    else
        return quantity;
}

int give_over(int quantity, int target_num) {
    if (target_num != 0)
        return (quantity - left_over(quantity, target_num)) / target_num;
    else
        return 0;
}

Prob 3. Friday the Thirteenth

This is a easy problem. Just match the days and $mod$ 7 to check and count the times.

/*
ID:
LANG: C
PROG: friday
 */

#include <stdio.h>
#include <stdlib.h>

int month_length(int year, int month);

int main() {
    FILE *fin = fopen("friday.in", "r");
    FILE *fout = fopen("friday.out", "w");
    int years;
    fscanf(fin, "%d", &years);
    int counter[7];
    for (int i = 0; i < 7; i++)
        counter[i] = 0;
    int current_year = 1900;
    int current_weekday = 1;
    for (int i = 0; i < years; i++) {
        for (int j = 1; j <= 12; j++) {
            counter[(current_weekday + 12) % 7]++;
            current_weekday += month_length(current_year, j);
            current_weekday %= 7;
        }
        current_year++;
    }
    fprintf(fout, "%d ", counter[6]);
    for (int i = 0; i < 5; i++)
        fprintf(fout, "%d ", counter[i]);
    fprintf(fout, "%d\n", counter[5]);
    exit(0);
}

int month_length(int year, int month) {
    int origin_length[12] =
        {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
        origin_length[1] += 1;
    return origin_length[month - 1];
}