CODEFORCE 672 题解（ABC）

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ROUND 672 题解(ABC)

A题题目

输入一个N 求第N个字符简单的字符串问题直接贴代码

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <complex>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <cassert>using namespace std;#define maxn 100005#define ll long long#define inf 2147483647#define INF 9223372036854775807#define pi acos(-1)const double eps=1e-9;string s;string l;void pre(){    for(int i=1;i<=1000;i++)    {        int k=i;        l="";        while(k>0)        {        l+=k%10+'0';        k/=10;        }        for(int j = l.length()-1;j>=0;j--)        {            s+=l[j];        }    }}int main(){    pre();    int n;    while(cin >> n)    cout << s[n-1] << endl;}

B题题目

字符串问题给你一个长度为n的字符串问你改变几个字符能让所有字串全不相同
考虑极限情况即可所有的单个字母都为字串只需要统计每个一个字母的个数

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <complex>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <cassert>using namespace std;#define maxn 100005#define ll long long#define inf 2147483647#define INF 9223372036854775807#define pi acos(-1)const double eps=1e-9;string s;string l;char a[maxn];int main(){    int n,i;    while(cin >> n)    {        getchar();        gets(a);        int sum= 1;        if(n>26)        cout << -1 << endl;        else        {            sort(a,a+n);            for(i=1;i<n;i++)            {                if(a[i-1]!=a[i])                    sum++;                    printf("%d",sum);            }             cout << n-sum << endl;        }    }}

C题题目

给出A B两人位置和垃圾桶位置同时还有N个垃圾的位置
问你拾完所有垃圾需要的最少路程
因为剪完垃圾都会回到垃圾桶的位置
所以只需要考虑第一次去捡垃圾的情况
先算出所有垃圾到垃圾桶的位置再乘2
然后再遍历求出 A B到垃圾的位置与垃圾桶位置的差值
再用总路程减去即可
第一次拾垃圾可以分为三种情况
1、A B 都有离垃圾近的这种情况并且不是离同一个垃圾
2、A B 都有离垃圾近的这种情况并且是同一个垃圾
3、只有A或者B有离垃圾近的情况
4、A B都比垃圾桶离垃圾远
1 这种情况应该算出一个离垃圾最短值和次短值然后用 A的最短值+B的次短值和B的最短值+A的次短值比较大小取较小的
2 即A加B最短值
3则让又最短值那个一直是垃圾另外一个永远休息
4 取比较不远那个一直拾垃圾另外一个休息
代码如下

#include <vector>#include <list>#include <map>#include <set>#include <deque>#include <queue>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <complex>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <cassert>using namespace std;#define maxn 100005#define ll long long#define inf 2147483647#define INF 9223372036854775807#define db doubleint i,m,n;struct pos {    int x,y;}a,b,t,c[maxn];double dis(pos xx,pos yy){    double dist = 0;    dist = (db(xx.x-yy.x)*(xx.x-yy.x)+(db)(xx.y-yy.y)*(xx.y-yy.y));    dist = sqrt(dist);    return dist;}int main(){    while(cin >>a.x >> a.y >> b.x >> b.y >>t.x >>t.y)    {        cin >> n;        db sum=0;        for(i=1;i<=n;i++)        {            cin >> c[i].x >> c[i].y;            sum +=2* dis(c[i],t);        }        db maa=0;        db mab=0;        double daa,dbb;        double ma=1e20;        db mb=1e20 ;        int flaga=0;        int flagb=0;        for(i=1;i<=n;i++)        {            daa = dis(a,c[i])-dis(c[i],t);            dbb = dis(b,c[i])-dis(c[i],t);            if(daa<maa)            {                ma = maa;                maa = daa;                flaga = i;            }            else if(daa < ma)            {                ma = daa;            }            if(dbb < mab)            {                mb = mab;                mab = dbb;                flagb = i;            }            else if(dbb < mb)            {                mb = dbb;            }        }        double temp = 0;        if(flaga&&flagb)        {            if(flaga==flagb)            {                temp+=min(maa+mb,mab+ma);            }            else            {                temp+=maa;                temp+=mab;            }        }        else if(flaga>0||flagb>0)        {            if(flaga>0)            {                temp+=maa;            }            if(flagb>0)            {                temp+=mab;            }        }        else        {            temp+=min(ma,mb);        }        printf("%.12lf\n",sum+temp);    }}

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