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[OJ 練習] 10300、102

テリ君(桃夫模式) | 2022-11-23 13:49:46 | 巴幣 2 | 人氣 194

10300 (1/5)

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

int main(){
    int m, n[20];
    int a[100][100], b[100][100], c[100][100];
    int money[100];

    scanf("%d", &m);
    if(m > 20) return 0;

    for(int i = 0; i < m; i++){
        scanf("%d", &n[i]);
        money[i] = 0;
        for(int j = 0; j < n[i]; j++){
            scanf("%d %d %d", &a[i][j], &b[i][j], &c[i][j]);
            money[i] = money[i] + a[i][j] * c[i][j];
        }
    }

    for(int i = 0; i < m; i++){
        printf("%d\n", money[i]);
    }

    return 0;
}

102 ((3)1/5)

#include <stdio.h>

int main(){
    int b[3], c[3], g[3];
    int min, result[6];
    int res_num;
    for(int j = 0; j < 2; j++){
        min = 0;
        for(int i = 0; i < 3; i ++){
            scanf("%d %d %d", &b[i], &g[i], &c[i]);
        }
        result[0] = b[1] + b[2] + c[0] + c[2] + g[0] + g[1];
        result[1] = b[1] + b[2] + c[0] + c[1] + g[0] + g[2];
        result[2] = b[0] + b[2] + c[1] + c[2] + g[0] + g[1];
        result[3] = b[0] + b[2] + c[0] + c[1] + g[1] + g[2];
        result[4] = b[0] + b[1] + c[1] + c[2] + g[0] + g[2];
        result[5] = b[0] + b[1] + c[0] + c[2] + g[1] + g[2];

        min = result[0];
        res_num = 0;
        for(int i = 0; i < 6; i ++){
            if(min > result[i]){
                min = result[i];
                res_num = i;
            }
        }
        if(res_num == 0){
            printf("BCG %d\n", min);
        }
        else if(res_num == 1){
            printf("BGC %d\n", min);
        }
        else if(res_num == 2){
            printf("CBG %d\n", min);
        }
        else if(res_num == 3){
            printf("GBC %d\n", min);
        }
        else if(res_num == 4){
            printf("CGB %d\n", min);
        }
        else if(res_num == 5){
            printf("GCB %d\n", min);
        }
        else return 1;
    }


    return 0;
}

102 會有個(3)是因為我基本上是用土法煉鋼法去做的

我覺得這樣做超沒效率但也只想得到這樣

需要直接爬文看看答案或是高人指點了TVT