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PAT乙级 1032 挖掘机技术哪家强 (20分) & 1033 旧键盘打字 (20分)
阅读量:610 次
发布时间:2019-03-12

本文共 1445 字,大约阅读时间需要 4 分钟。

1032 挖掘机技术哪家强 (20分)

只需要注意数组个数要取大,不然最后一个测试点出错。

#include
#include
#include
using namespace std;int main() { int sum = 0; cin >> sum; int bian = 0, fen = 0,max=0,maxbian=0,maxfen=0; int arr[1000000] = { 0 };//这里一定要取大,不然最后一个测试点不通过 for (int i = 0; i < sum; i++) { cin >> bian >> fen; arr[bian] += fen; if (bian > max)//找出最大编号,减少计算次数 max = bian; } for (int i = 1; i <= max; i++)//找出最大点 { if (maxfen < arr[i]) { maxbian = i; maxfen = arr[i]; } } cout << maxbian << " " << maxfen; return 0;}

1033 旧键盘打字 (20分)

需要用getline输入,因为题目要求是坏键盘可以为空,cin无法输入空,最后一个测试点就是过不去????

#include
#include
#include
using namespace std;int main(){ string a, b; getline(cin, a); getline(cin, b); int temp = 0; if (!a.size()) { for (int i = 0; i < b.size(); i++) cout << b[i]; return 0; } for (int i = 0; i < a.size(); i++) { if (a[i] == '+') { for (int i = 0; i < b.size(); i++) { if (b[i] >= 'A' && b[i] <= 'Z') b[i] = '*'; if (b[i] == '+') b[i] = '*'; } } else if (a[i] >= 'A' && a[i] <= 'Z') { char temp = a[i] - 'A' + 'a'; for (int i1 = 0; i1 < b.size(); i1++) { if(b[i1]==temp) b[i1] = '*'; } } else { for (int i1 = 0; i1 < b.size(); i1++) { if(b[i1]==a[i]) b[i1] = '*'; } } } for (int i = 0; i < b.length(); i++) { if (b[i] != '*') { cout << b[i]; temp = 1; } } if (temp == 0)//如果全都是无法输出就输出一个回车 { cout << endl; } return 0;}

 

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