pre过了三题 后来A题被hack了 B题终测挂了 两题其实都是有一个小细节没有处理好 中间C还因为cinT了一次
唉本来打的还不错的 还是太菜了 继续加油吧
A-Benches
有n张椅子 原来第i张上有ai个人 现在来了m个人
求最大值的最大和最小的可能
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<set> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<vector> 9 #include<queue> 10 #include<stack> 11 12 using namespace std; 13 14 int n, m; 15 const int maxn = 105; 16 int a[maxn]; 17 18 int main() 19 { 20 while (cin >> n >> m) { 21 int maxpeo = -1; 22 for (int i = 0; i < n; i++) { 23 cin >> a[i]; 24 maxpeo = max(maxpeo, a[i]); 25 } 26 27 int maxk = maxpeo + m; 28 int sub = 0; 29 for (int i = 0; i < n; i++) { 30 sub += maxpeo - a[i]; 31 } 32 int mink; 33 if(sub >= m){ 34 mink = maxpeo; 35 } 36 else if ((m - sub) % n == 0) { 37 mink = (m - sub) / n + maxpeo; 38 } 39 else { 40 mink = (m - sub) / n + 1 + maxpeo; 41 } 42 43 cout<<mink<<" "<<maxk<<endl; 44 } 45 }
B-Vitamins
有n瓶果汁，每瓶有一个价格 和 可能含有的维生素【最多只有ABC三种维生素】
现在想要用最少的钱集齐ABC
本来感觉是dp 写不出 所以先去写了C 后来发现其实贪心就能过
分类讨论一下 一种是 直接选三个维生素都有的里面价格最小的
一种是 选两个维生素加缺的里面价格最小的
一种是 选单个维生素价格最小的之和
存好含三种的果汁 两种的果汁 含A的果汁【不包括三种都有】含B的含C的
第一种可能只有一种情况
第二种可能 枚举 跑一遍
第三种可能也只有一种情况
比一下最小值
需要注意有可能有的数量是0
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<set> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<vector> 9 #include<queue> 10 #include<stack> 11 12 #define inf 0x3f3f3f3f 13 14 using namespace std; 15 16 const int maxn = 1005; 17 int n; 18 struct node { 19 int c; 20 char vitamin[5]; 21 }juice[maxn]; 22 node a[maxn], b[maxn], c[maxn], three[maxn], two[maxn]; 23 24 bool cmp(node a, node b) 25 { 26 return a.c < b.c; 27 } 28 29 int main() 30 { 31 while (scanf("%d", &n) != EOF) { 32 int cnta = 0, cntb = 0, cntc = 0, cntth = 0, cnttwo = 0; 33 for (int i = 0; i < n; i++) { 34 scanf("%d %s", &juice[i].c, juice[i].vitamin); 35 if (strlen(juice[i].vitamin) == 3) { 36 three[cntth++] = juice[i]; 37 } 38 else if (strlen(juice[i].vitamin) == 2) { 39 two[cnttwo++] = juice[i]; 40 for (int j = 0; j < 2; j++) { 41 if (juice[i].vitamin[j] == 'A') { 42 a[cnta++] = juice[i]; 43 } 44 else if (juice[i].vitamin[j] == 'B') { 45 b[cntb++] = juice[i]; 46 } 47 else { 48 c[cntc++] = juice[i]; 49 } 50 } 51 } 52 else { 53 if (juice[i].vitamin[0] == 'A') { 54 a[cnta++] = juice[i]; 55 } 56 else if (juice[i].vitamin[0] == 'B') { 57 b[cntb++] = juice[i]; 58 } 59 else { 60 c[cntc++] = juice[i]; 61 } 62 } 63 } 64 65 if (cntth == 0 && (cnta == 0 || cntb == 0 || cntc == 0)) { 66 printf("-1 "); 67 continue; 68 } 69 70 sort(three, three + cntth, cmp); 71 sort(two, two + cnttwo, cmp); 72 sort(a, a + cnta, cmp); 73 sort(b, b + cntb, cmp); 74 sort(c, c + cntc, cmp); 75 76 int ans; 77 if(cntth == 0){ 78 ans = inf; 79 } 80 else{ 81 ans = three[0].c; 82 } 83 for (int i = 0; i < cnttwo; i++) { 84 if (strcmp(two[i].vitamin, "AB") == 0 || strcmp(two[i].vitamin, "BA") == 0) { 85 ans = min(ans, two[i].c + c[0].c); 86 } 87 else if (strcmp(two[i].vitamin, "AC") == 0 || strcmp(two[i].vitamin, "CA") == 0) { 88 ans = min(ans, two[i].c + b[0].c); 89 } 90 else { 91 ans = min(ans, two[i].c + a[0].c); 92 } 93 } 94 if(cnta != 0 && cntb != 0 && cntc != 0){ 95 ans = min(ans, a[0].c + b[0].c + c[0].c); 96 } 97 98 99 printf("%d ", ans); 100 } 101 }
C-Array Product
给一组数列 做n-1次操作
有两种操作 ：第一种将ai和aj相乘结果赋给aj，ai删除
第二种 将ai位置删除【这种操作最多只能一次】
其实也是贪心
首先 所有的0都要合并 然后删去
负数 如果是奇数个的话 删去绝对值最小的那个
如果既有0又有奇数个负数 就把这个负数和0合并再删去
剩下的这些全部照常做第一种操作就好了
1 #include <iostream> 2 #include<algorithm> 3 #include<stdio.h> 4 #include<set> 5 #include<cmath> 6 #include<cstring> 7 #include<map> 8 #include<vector> 9 #include<queue> 10 #include<stack> 11 12 #define inf 0x3f3f3f3f 13 14 using namespace std; 15 16 int n; 17 const int maxn = 2e5 + 5; 18 int a[maxn]; 19 bool vis[maxn]; 20 21 int main() 22 { 23 //freopen("C:\Users\wyb\Desktop\tmpcode\codeforces\in.txt", "r", stdin); 24 //freopen("C:\Users\wyb\Desktop\tmpcode\codeforces\out.txt", "w", stdout); 25 while (scanf("%d", &n) != EOF) { 26 memset(vis, 1, sizeof(vis)); 27 int cntmin = 0, lastzero = -1, maxminpos, maxmin = -inf, cnt = 0; 28 for (int i = 1; i <= n; i++) { 29 scanf("%d", &a[i]); 30 if (a[i] == 0) { 31 if (lastzero == -1) { 32 lastzero = i; 33 } 34 else { 35 printf("1 %d %d ", lastzero, i); 36 //cout << "1 " << lastzero << " " << i << endl; 37 vis[lastzero] = false; 38 lastzero = i; 39 cnt++; 40 } 41 } 42 if (a[i] < 0) { 43 cntmin++; 44 if (maxmin < a[i]) { 45 maxmin = a[i]; 46 maxminpos = i; 47 } 48 } 49 } 50 if (cntmin % 2 && lastzero != -1) { 51 if(cnt == n - 1){ 52 continue; 53 } 54 cnt++; 55 printf("1 %d %d ", maxminpos, lastzero); 56 //cout << "1 " << maxminpos << " " << lastzero << endl; 57 if(cnt == n - 1){ 58 continue; 59 } 60 cnt++; 61 printf("2 %d ", lastzero); 62 //cout << "2 " << lastzero << endl; 63 vis[maxminpos] = vis[lastzero] = false; 64 } 65 else if (lastzero != -1) { 66 if(cnt == n - 1){ 67 continue; 68 } 69 cnt++; 70 printf("2 %d ", lastzero); 71 //cout << "2 " << lastzero << endl; 72 vis[lastzero] = false; 73 } 74 else if(cntmin % 2){ 75 if(cnt == n - 1){ 76 continue; 77 } 78 cnt++; 79 printf("2 %d ", maxminpos); 80 //cout<<"2 "<<maxminpos<<endl; 81 vis[maxminpos] = false; 82 } 83 84 int last = -1; 85 for (int i = 1; i <= n; i++) { 86 if (vis[i]) { 87 if (last == -1) { 88 last = i; 89 } 90 else { 91 if(cnt == n - 1){ 92 break; 93 } 94 cnt++; 95 printf("1 %d %d ", last, i); 96 //cout << "1 " << last << " " << i << endl; 97 last = i; 98 } 99 } 100 } 101 //cout<<endl; 102 } 103 }