codeforces#510 Div2

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

 #include <iostream>
 #include<algorithm>
 #include<stdio.h>
 #include<set>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<vector>
 #include<queue>
 #include<stack>

 using namespace std;

 int n, m;
 const int maxn = ;
 int a[maxn];

 int main()
 {
     while (cin >> n >> m) {
         int maxpeo = -;
         for (int i = ; i < n; i++) {
             cin >> a[i];
             maxpeo = max(maxpeo, a[i]);
         }

         int maxk = maxpeo + m;
         int sub = ;
         for (int i = ; i < n; i++) {
             sub += maxpeo - a[i];
         }
         int mink;
         if(sub >= m){
             mink = maxpeo;
         }
         else if ((m - sub) % n == ) {
             mink = (m - sub) / n + maxpeo;
         }
         else {
             mink = (m - sub) / n +  + maxpeo;
         }

         cout<<mink<<" "<<maxk<<endl;
     }
 }

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

 #include <iostream>
 #include<algorithm>
 #include<stdio.h>
 #include<set>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<vector>
 #include<queue>
 #include<stack>

 #define inf 0x3f3f3f3f

 using namespace std;

 const int maxn = ;
 int n;
 struct node {
     int c;
     char vitamin[];
 }juice[maxn];
 node a[maxn], b[maxn], c[maxn], three[maxn], two[maxn];

 bool cmp(node a, node b)
 {
     return a.c < b.c;
 }

 int main()
 {
     while (scanf("%d", &n) != EOF) {
         int cnta = , cntb = , cntc = , cntth = , cnttwo = ;
         for (int i = ; i < n; i++) {
             scanf("%d %s", &juice[i].c, juice[i].vitamin);
             if (strlen(juice[i].vitamin) == ) {
                 three[cntth++] = juice[i];
             }
             else if (strlen(juice[i].vitamin) == ) {
                 two[cnttwo++] = juice[i];
                 for (int j = ; j < ; j++) {
                     if (juice[i].vitamin[j] == 'A') {
                         a[cnta++] = juice[i];
                     }
                     else if (juice[i].vitamin[j] == 'B') {
                         b[cntb++] = juice[i];
                     }
                     else {
                         c[cntc++] = juice[i];
                     }
                 }
             }
             else {
                 if (juice[i].vitamin[] == 'A') {
                     a[cnta++] = juice[i];
                 }
                 else if (juice[i].vitamin[] == 'B') {
                     b[cntb++] = juice[i];
                 }
                 else {
                     c[cntc++] = juice[i];
                 }
             }
         }

         if (cntth ==  && (cnta ==  || cntb ==  || cntc == )) {
             printf("-1\n");
             continue;
         }

         sort(three, three + cntth, cmp);
         sort(two, two + cnttwo, cmp);
         sort(a, a + cnta, cmp);
         sort(b, b + cntb, cmp);
         sort(c, c + cntc, cmp);

         int ans;
         if(cntth == ){
             ans = inf;
         }
         else{
             ans = three[].c;
         }
         for (int i = ; i < cnttwo; i++) {
             if (strcmp(two[i].vitamin, "AB") ==  || strcmp(two[i].vitamin, "BA") == ) {
                 ans = min(ans, two[i].c + c[].c);
             }
             else if (strcmp(two[i].vitamin, "AC") ==  || strcmp(two[i].vitamin, "CA") == ) {
                 ans = min(ans, two[i].c + b[].c);
             }
             else {
                 ans = min(ans, two[i].c + a[].c);
             }
         }
         if(cnta !=  && cntb !=  && cntc != ){
             ans = min(ans, a[].c + b[].c + c[].c);
         }

         printf("%d\n", ans);
     }
 }

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

 #include <iostream>
 #include<algorithm>
 #include<stdio.h>
 #include<set>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<vector>
 #include<queue>
 #include<stack>

 #define inf 0x3f3f3f3f

 using namespace std;

 int n;
 const int maxn = 2e5 + ;
 int a[maxn];
 bool vis[maxn];

 int main()
 {
     //freopen("C:\\Users\\wyb\\Desktop\\tmpcode\\codeforces\\in.txt", "r", stdin);
     //freopen("C:\\Users\\wyb\\Desktop\\tmpcode\\codeforces\\out.txt", "w", stdout);
     while (scanf("%d", &n) != EOF) {
         memset(vis, , sizeof(vis));
         int cntmin = , lastzero = -, maxminpos, maxmin = -inf, cnt = ;
         for (int i = ; i <= n; i++) {
             scanf("%d", &a[i]);
             if (a[i] == ) {
                 if (lastzero == -) {
                     lastzero = i;
                 }
                 else {
                     printf("1 %d %d\n", lastzero, i);
                     //cout << "1 " << lastzero << " " << i << endl;
                     vis[lastzero] = false;
                     lastzero = i;
                     cnt++;
                 }
             }
             if (a[i] < ) {
                 cntmin++;
                 if (maxmin < a[i]) {
                     maxmin = a[i];
                     maxminpos = i;
                 }
             }
         }
         if (cntmin %  && lastzero != -) {
             if(cnt == n - ){
                 continue;
             }
             cnt++;
             printf("1 %d %d\n", maxminpos, lastzero);
             //cout << "1 " << maxminpos << " " << lastzero << endl;
             if(cnt == n - ){
                 continue;
             }
             cnt++;
             printf("2 %d\n", lastzero);
             //cout << "2 " << lastzero << endl;
             vis[maxminpos] = vis[lastzero] = false;
         }
         else if (lastzero != -) {
             if(cnt == n - ){
                 continue;
             }
             cnt++;
             printf("2 %d\n", lastzero);
             //cout << "2 " << lastzero << endl;
             vis[lastzero] = false;
         }
         else if(cntmin % ){
             if(cnt == n - ){
                 continue;
             }
             cnt++;
             printf("2 %d\n", maxminpos);
             //cout<<"2 "<<maxminpos<<endl;
             vis[maxminpos] = false;
         }

         int last = -;
         for (int i = ; i <= n; i++) {
             if (vis[i]) {
                 if (last == -) {
                     last = i;
                 }
                 else {
                     if(cnt == n - ){
                         break;
                     }
                     cnt++;
                     printf("1 %d %d\n", last, i);
                     //cout << "1 " << last << " " << i << endl;
                     last = i;
                 }
             }
         }
         //cout<<endl;
     }
 }