1 条题解

  • 0
    @ 2024-12-24 9:59:27

    C :

    #include <stdio.h>
    #include <string.h>
    
    int main(){
    	
    	int m, s, c, i, j, k, number, tmp, flag, min, max, result,
    		goods[256], notOccupied[256];
    	
    	while (EOF != scanf("%d%d%d", &m, &s, &c)){
    		
    		min = 256;
    		max = 0;
    		memset(goods, 0, sizeof(goods));
    		while (c--){
    			
    			scanf("%d", &tmp);
    			
    			goods[tmp] = 1;
    			if (tmp > max){
    				max = tmp;
    			}
    			if (tmp < min){
    				min = tmp;
    			}
    		}
    		
    		k = 0;
    		tmp = 0;
    		flag = 0;
    		for (i = min; i <= max; i++){
    			
    			if (goods[i]){
    				if (flag){
    					notOccupied[k++] = tmp;
    				}
    				tmp = 0;
    				flag = 0;
    				continue;
    			}
    			
    			tmp++;
    			flag = 1;
    		}
    		
    		for (i = 0; i < k; i++){
    			
    			for (j = i + 1; j < k; j++){
    				
    				if (notOccupied[i] < notOccupied[j]){
    					tmp = notOccupied[i];
    					notOccupied[i] = notOccupied[j];
    					notOccupied[j] = tmp;
    				}
    			}
    		}
    		
    		result = max - min + 1;
    		
    		for (i = 0; i < m - 1 && i < k; i++){
    			
    			result -= notOccupied[i];
    		}
    		
    		printf("%d\n", result);
    	}
    	
    	return 0;
    }
    
    

    C++ :

    #include <cstdio>
    #include <algorithm>
    #include <cstring>
    using namespace std;
    const int V = 200 + 5;
    int m, s, c, len[V], ans[V];
    int main() {
        int i, j;
        while(~scanf("%d%d%d", &m, &s, &c)) {
            for(i = 0; i < c; ++i)
                scanf("%d", &len[i]);
            sort(len, len + c);
            for(i = 1; i < c; ++i)
                ans[i - 1] = len[i] - len[i - 1] - 1;
            sort(ans, ans + c - 1);
            int L = len[c - 1] - len[0] + 1;
            while(--m && c--)
                L -= ans[c - 1];
            printf("%d\n", L);
        }
    }
    
    • 1

    信息

    ID
    1514
    时间
    1000ms
    内存
    128MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者