A1048 Find Coins （25 points）hash/二分/two pointers

A1048 Find Coins

哈希写法

• 硬币面值有限且为整数，可直接作为下标，计数即可。

#include 

using namespace std;

int main() {
    int coins[1001] = {0}, nn, total, value;
    scanf("%d%d", &nn, &total);
    for (int i = 0; i < nn; ++i) {
        scanf("%d", &value);
        coins[value]++;
    }
    bool ok = false;
    for (int i = 1; i <= total / 2; ++i) {
        if (coins[i]) {
            if (coins[total - i]) {
                if (total - i != i || (total - i == i && coins[i] >= 2)) {
                    printf("%d %d\n", i, total - i);
                    ok = true;
                    break;
                }
            }
        }
    }
    if (!ok)
        puts("No Solution");
    return 0;
}

排序➕二分

#include 
#include 

using namespace std;

int main() {
    int coins[100010], nn, total, c1 = 0, c2 = 0;
    scanf("%d%d", &nn, &total);
    for (int i = 0; i < nn; i++)
        scanf("%d", &coins[i]);
    sort(coins, coins + nn);
    for (int i = 0; i < nn - 1; ++i) {
        int target = total - coins[i];
        int index = lower_bound(coins + i + 1, coins + nn, target) - coins;
        if (coins[index] == target) {
            c1 = coins[i], c2 = coins[index];
            break;
        }
    }
    if (c1 == 0)
        puts("No Solution");
    else printf("%d %d\n", c1, c2);
    return 0;
}

two pointers

排序，下标left from 0，right from n-1
若大，right--; 若小，left++; 若等，break。
当left＞right，结束。