ct1807部分题的题解

Wang Haiyang is a strong and optimistic Chinese youngster. Although born and brought up in the northern inland city Harbin, he has deep love and yearns for the boundless oceans. After graduation, he came to a coastal city and got a job in a marine transportation company. There, he held a position as a navigator in a freighter and began his new life.

The cargo vessel, Wang Haiyang worked on, sails among 6 ports between which exist 9 routes. At the first sight of his navigation chart, the 6 ports and 9 routes on it reminded him of Graph Theory that he studied in class at university. In the way that Leonhard Euler solved The Seven Bridges of Königsberg, Wang Haiyang regarded the navigation chart as a graph of Graph Theory. He considered the 6 ports as 6 nodes and 9 routes as 9 edges of the graph. The graph is illustrated as below.

According to Graph Theory, the number of edges related to a node is defined as Degree number of this node.

Wang Haiyang looked at the graph and thought, “If arranged, the Degree numbers of all nodes of graph G can form such a sequence: 4, 4, 3,3,2,2, which is called the degree sequence of the graph. Of course, the degree sequence of any simple graph (according to Graph Theory, a graph without any parallel edge or ring is a simple graph) is a non-negative integer sequence”

Wang Haiyang is a thoughtful person and tends to think deeply over any scientific problem that grabs his interest. So as usual, he also gave this problem further thought, “as we know, any a simple graph always corresponds with a non-negative integer sequence. But whether a non-negative integer sequence always corresponds with the degree sequence of a simple graph? That is, if given a non-negative integer sequence, are we sure that we can draw a simple graph according to it.”

Let’s put forward such a definition: provided that a non-negative integer sequence is the degree sequence of a graph without any parallel edge or  ring, that is, a simple graph, the sequence is draw-possible,  otherwise, non-draw-possible. Now the problem faced with Wang Haiyang is how to test whether a non-negative integer sequence is draw-possible or not. Since Wang Haiyang hasn’t studied Algorithm Design course, it is difficult for him to solve such a problem. Can you help him?

输入

The first line of input contains an integer T, indicates the number of test cases. In each case, there are n+1 numbers; first is an integer n (n<1000), which indicates there are n integers in the sequence; then follow n integers, which indicate the numbers of the degree sequence.

输出

For each case, the answer should be “yes” or “no”, indicating this case is “draw-possible” or “non-draw-possible”.

这里考察了判定简单图专用算法：Havel算法。算法步骤如下：

1. 从大到小排序
2. 删除第一个数。并且“删数后的第一个数至第一个数数值”范围内的所有数全部减一。
3. 重复这个操作，如果出现某个数字大于n-i（当前点的个数）或者某个数小于零，这都不成立。

//#include<pch.h>
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#include<queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define DETERMINATION main
#define lldin(a) scanf_s("%lld", &a)
#define println(a) printf("%lld\n", a)
#define reset(a, b) memset(a, b, sizeof(a))
#define debug cout<<"procedures above are available"<<endl;
using namespace std;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int mod = 1000000007;
const int tool_const = 19991126;
const int tool_const2 = 33;
inline ll nextlong()
{
    ll tmp = 0, si = 1;
    char c;
    c = getchar();
    while (c > '9' || c < '0')
    {
        if (c == '-')
            si = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        tmp = tmp * 10 + c - '0';
        c = getchar();
    }
    return si * tmp;
}
/**Maintain your determination.Nobody knows the magnificent landscape
at his destination before the arrival with stumble.**/
/**Last Remote**/
ll a[500000];
bool cmp(ll a,ll b)
{
    return a>b;
}
int DETERMINATION()
{
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n;
        cin>>n;
        reset(a,0);
        for(int i=1;i<=n;i++)
            cin>>a[i];
        bool sign=true;
        for(int i=1;i<=n&&sign;i++)
        {
            sort(a+1+i,a+n+1,cmp);
            ll limit=a[i];
            if(limit>n-i)//超出最大限制
                sign=false;
            for(int j=1;j<=limit&&sign;j++)
            {
                if(a[i+j]<=0)//负数
                    sign=false;
                a[i+j]--;
            }
        }
        if(sign)
            cout<<"yes"<<endl;
        else
            cout<<"no"<<endl;
    }
    return 0;
}
 

 

A luxury yacht with 100 passengers on board is sailing on the sea in the twilight. The yacht  is ablaze with lights and there comes out laughers and singing from the hall where an evening party is in full swing. People are singing, dancing and enjoying themselves.

The yacht is equipped with the most advanced navigation and driving system which can all be manipulated by a computer. When the captain notices that there is only gentle breeze and the sea waves are not high, he starts the autopilot. The yacht sails forward smoothly, ploughs the waves. When it’s completely dark, the passengers start to feel a little funny for sudden forward rushes or sudden decelerations or slight swings. The captain immediately walks to the driving platform and switches the autopilot to human manipulation. The yacht returns back to normal and the party restarts. Laughers come back, too.

The captain summons the engineer on board to do a thorough check of the navigation system. It turns out that only the computer is out of order, but the exact failure is still unclear. There is a computer scientist among the passengers who is also invited to the cab to give a hand. He first inputs several groups of data to test the computer. When he inputs 1+2+3, the computer outputs 6, which is exactly right. But when he inputs 4+5+6, the computer outputs 5, which is wrong. Then he inputs 12+13+14, and gets 39, another right answer, while he inputs 14+15+16, and gets 35, another wrong answer. After the test, the computer scientist says smilingly: “the failure is clear now. The computer's adder can not carry." After excluding the failure, the captain restarts the autopilot and the yacht returns back to normal, sailing smoothly on the sea.

The captain and the engineer invite the computer scientist to sit down and have a talk. The computer scientist tells a story as following:

A former mathematician defined a kind of simple addition expression. 
If there is an expression (i) + (i+1) + (i+2), i>=0, when carried out additive operations, no position has a carry, it is called simple addition expression.

For instance, when i equals 0, 0+1+2 is a simple addition expression, meanwhile when i equals 11, 11+12+13 is a simple addition expression, too. Because of that no position has a carry.

However, when i equals 3, 3+4+5 is not a simple addition expression, that is because 3+4+5 equals 12, there is a carried number from unit digit to tens digit. In the same way, when i equals 13, 13+14+15 is not a simple addition expression, either. However, when i equals 112, 112+113+114 is a simple addition expression. Because 112+113+114 equals 339, there is no carry in the process of adding.

when the students have got the definition of simple addition expression, the mathematician puts forward a new question: for a positive integer n, how many simple addition expressions exist when i<n. In addition, i is the first number of a simple addition expression.

when the value of n is large enough, the problem needs to be solved by means of computer.

简单的数位计算。

//#include<pch.h>
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
#include<queue>
#include <map>
#include <algorithm>
#include <stack>
#include <iomanip>
#include <cstring>
#include <cmath>
#define DETERMINATION main
#define lldin(a) scanf_s("%lld", &a)
#define println(a) printf("%lld\n", a)
#define reset(a, b) memset(a, b, sizeof(a))
#define debug cout<<"procedures above are available"<<endl;
using namespace std;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int mod = 1000000007;
const int tool_const = 19991126;
const int tool_const2 = 33;
inline ll nextlong()
{
    ll tmp = 0, si = 1;
    char c;
    c = getchar();
    while (c > '9' || c < '0')
    {
        if (c == '-')
            si = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        tmp = tmp * 10 + c - '0';
        c = getchar();
    }
    return si * tmp;
}
/**Maintain your determination.Nobody knows the magnificent landscape
at his destination before the arrival with stumble.**/
/**Last Remote**/
ll digits[50];
ll cnt = 0;
void div2(ll x)
{
    while (x)
    {
        digits[++cnt] = x % 10;
        x /= 10;
    }
}
ll slowpower(ll a,ll b)
{
    ll ans = 1;
    for (int i = 1; i <= b; i++)
        ans *= a;
    return ans;
}
int DETERMINATION()
{
    ///ios::sync_with_stdio(false);
    ///cin.tie(0), cout.tie(0);
    ll n;
    while (~scanf("%lld",&n))
    {
        reset(digits, 0);
        cnt = 0;
        div2(n);
        ll ans = 0;
        bool sign = false;
        for (int i = cnt; i >= 2; i--)
        {
            if (digits[i] > 3)//如果大于三，就意味着超出了最大范围，可以加上所有的值
            {
                sign = true;
                ans += 3 * slowpower(4, max(i-1,0));
                break;
            }
            else
                ans += digits[i] * 3 * slowpower(4, max(i-2,0));
        }
        if (!sign)
        {
            if (digits[1] > 3)
                ans += 3;
            else
                ans += digits[1];
        }
        println(ans);
    }
    return 0;
}