This game is a good ghost and beast. A question has broken down wa four times. The mentality has exploded directly. I wanted to give up treatment, but I found that BCD is a silly question.

A. If at first you don't succeed... (Principle of inclusion and exclusion)

Main idea of the title:

Some $N individuals took the exam. Among those who passed the exam, some $A individuals went to the first hotel party, some $B individuals went to the second hotel party, and some $C individuals went to two hotel parties at the same time.

Ask how many people failed the exam (the protagonist failed the exam)

Sol

The pupils scolded it. There must be $A+B-C $individuals at the party. If you subtract the amount from $N, it's unqualified.

At first wa exploded four times and then added a special sentence A.

#include<cstdio>
using namespace std;
int main() {
    int A, B, C, N;
    scanf("%d %d %d %d", &A, &B, &C, &N);
    int no = N - (A + B - C);
    if(A >= N || B >= N || C >= N || (C > A) || (C > B)) {printf("-1"); return 0;}
    if(no <= 0 || no > N) {printf("-1"); return 0;}
    printf("%d", N - (A + B - C));
    return 0;
}

B. Getting an A (Tan Xin)

Main idea of the title:

Vasya did the $n $door experiment and scored between $2 and $5. He wanted to round his average score (total score divided by the number of experiments) to $5. He asked for at least a few more experiments.

Sol

Directly greedy, it must be done a little again first, first sorting, enumeration to determine whether it has been qualified on the line.

#include<cstdio>
#include<algorithm>
using namespace std;
const int MAXN = 1e5 + 10; 
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int a[MAXN]; 
int main() {
    double N = read(),  S = 0;
    int x = N;
    for(int i = 1; i <= N; i++) a[i] = read(), S += a[i];
    sort(a + 1, a + x + 1);
    if(S / N >= 4.5) {printf("0"); return 0;}
    for(int i = 1; i <= N; i++) {
        S -= a[i]; S += 5;
        if(S / N >= 4.5) {
            printf("%d", i); return 0;
        }
    }
    return 0;
}

C. Candies (two points)

Main idea of the title:

Vasya had $n $candy, and at the beginning Vasya chose an integer of $k $to indicate that he would eat $k $candy a day. Petya wanted to steal some candy, and he would eat the current amount of $10\$(down to the whole) candy every day.

Note: If Vasya should eat candy and is not satisfied with $k, Vasya will eat it all. If Petya eats less than $10 candy, then Petya won't eat candy.

Output the smallest $k $so that $Vasya $eats at least half of the candy

Sol

Obviously, $k $is monotonous. Divide it directly.

Because $Petya $eats $10\\\\ candy at a time, the process of eating candy is fast and can be simulated directly.

#include<cstdio>
#include<algorithm>
#define int long long 
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N;
bool check(int k) {
    int cur = 1, Ansa = 0, Ansb = 0, now = N;
    while(now > 0) {
        if(cur & 1) Ansa += min(k, now), now -= min(k, now);
        else Ansb += now / 10, now -= now / 10;
        cur ^= 1;
    }
    return Ansa >= Ansb;
}
main() {
    N = read();
    int l = 1, r = N, ans = 0;
    check(3);
    while(l <= r) {
        int mid = l + r >> 1;
        if(check(mid)) r = mid - 1, ans = mid;
        else l = mid + 1;
    }
    printf("%I64d", ans);
}

D. Bishwock(dp)

Main idea of the title:

Bishwock is a special kind of chess. Its placement rules are as follows (similar to Tetris)

Give an initial scenario of $2*n $and ask how many chess pieces you can play at most (X means you can't play them here, 0 means you can play them several times)

Sol

Ran1's code is amazing. You can't learn it.

I only violence dp, let $f[i][j] $denote to the line of $i $and the state is the maximum of $j $(four states), $g[i] $denotes the maximum of all States of the line of $i $(max for f[i][1/2/3/4).)

Judge where you can move from when you move.

 #include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int MAXN = 1001;
char s1[MAXN], s2[MAXN];
int f[MAXN][5], g[MAXN];
int getg(int x) {
    int ans = 0;
    for(int i = 1; i <= x; i++)
        ans = max(ans, g[i]);
    return ans;
}
int main() {
    scanf("%s %s", s1 + 1, s2 + 1);
    int N = strlen(s1 + 1), ans = 0;
    for(int i = 2; i <= N; i++) {
        for(int k = 1; k <= 4; k++) 
            f[i][k] = getg(i - 1);
        if(s1[i] == '0' && s2[i] == '0') {
            if(s1[i - 1] == '0') {
                f[i][2] = max(f[i][2], g[i - 2] + 1);
                if(s2[i - 1] == '0')
                    f[i][2] = max(f[i][2], f[i - 1][1] + 1);
            }
            if(s2[i - 1] == '0') {
                f[i][3] = max(f[i][3], g[i - 2] + 1);
                if(s1[i - 1] == '0')
                    f[i][3] = max(f[i][3], f[i - 1][4] + 1);
            }
        }
        if(s1[i] == '0') {
            if(s1[i - 1] == '0' && s2[i - 1] == '0') 
                f[i][4] = max(f[i][4], g[i - 2] + 1);
        } 
        if(s2[i] == '0') {
            if(s1[i - 1] == '0' && s2[i - 1] == '0')
                f[i][1] = max(f[i][1], g[i - 2] + 1);
        } 
        for(int k = 1; k <= 4; k++)    
            g[i] = max(f[i][k], g[i]);
        ans = max(ans, g[i]);
    }
    printf("%d", ans);
    return 0;
}

summary

The war was terrible.