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  • 其他

    暴力/练习

    https://codeforces.com/contest/1293/problem/D

    注意:(t leq 10^{16}),要将inf设置成(10^{17})

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int maxn = 1000;
    const ll limit = 1e17;
    ll x[maxn], y[maxn];
    
    int main() {
        ll  ax, bx, ay, by;
        ll xs, ys, t;
        cin >> x[0] >> y[0] >> ax >> ay >> bx >> by >> xs >> ys >> t;
        int cnt = 0;
        while(x[cnt]<(limit-bx+1)/ax&&y[cnt]<(limit-by+1)/ay) {
            cnt++;
            x[cnt] = x[cnt - 1] * ax + bx;
            y[cnt] = y[cnt - 1] * ay + by;
        }
        ll ans = 0;
        for (int i = 0; i <= cnt;i++) {
            for (int j = 0; j <= cnt;j++) {
                ll a = abs(xs - x[i]) + abs(ys - y[i]);
                ll b = abs(x[i] - x[j]) + abs(y[i] - y[j]);
                if(a<=t-b) {
                    ll m = abs(i - j) + 1;
                    ans = max(ans, m);
                }
            }
        }
        cout << ans << endl;
    }
    

    中缀表达式 Bracket Sequence

    https://nanti.jisuanke.com/t/43466

    就是算一下

    #include<bits/stdc++.h>
    using namespace std;
    const long long mod = 1e9 + 7;
    const int maxn = 3e5+10;
    vector<long long> v[maxn];
    int num(string s) {
        int len = s.size();
        int ret = 0;
        for (int i = 0; i < len;i++) {
            ret = ret * 10 + s[i] - '0';
        }
        return ret;
    }
    int main() {
        int n;
        scanf("%d", &n);
        int cnt = 0;
        for (int i = 1; i <= n;i++) {
            string s;
            cin >> s;
            if(s[0]=='(') {
                cnt++;
            } 
            else if(s[0]==')') {
                if(cnt%2==0) {
                    long long x = 0;
                    while(v[cnt].size()) {
                        x += v[cnt].back();
                        x %= mod;
                        v[cnt].pop_back();
                    }
                    cnt--;
                    v[cnt].push_back(x);
                }
                else {
                    long long x = 1;
                    while(v[cnt].size()) {
                        x *= v[cnt].back();
                        x %= mod;
                        v[cnt].pop_back();
                    }
                    cnt--;
                    v[cnt].push_back(x);
                }
            }
            else{
                int x = num(s);
                v[cnt].push_back(x);
            }
        }
    
        long long ans = 0;
        while(v[0].size()) {
            ans = ans + v[0].back();
            ans %= mod;
            v[0].pop_back();
        }
        printf("%lld
    ", ans);
    }
    

    [CF1175B] catch overflow

    #include<bits/stdc++.h>
    using namespace std;
    const long long limit = ((long long)1 << 32) - 1;
    
    int main() {
        int n;
        scanf("%d", &n);
        long long ans = 0,t = 1;
        int f = 0, cnt = 0;
        vector<long long> v;
        for (int i = 1; i <= n;i++) {
            string s;
            cin >> s;
            if(s[0]=='a') {
                ans += t;
                if(ans>limit) {
                    f = 1;
                }
            }
            if(s[0]=='f'){
                int x;
                scanf("%d", &x);
                if(t>limit){
                    cnt++;
                    continue;
                }
                t *= x;
                v.push_back(x);
            }
            if(s[0]=='e') {
                if(cnt==0){
                    t /= v.back();
                    v.pop_back();
                }
                else
                    cnt--;
            }
        }
        if(f)
            printf( "OVERFLOW!!!
    " );
    else
        cout << ans << endl;
    }
    

    NowCoder 非递归实现组合型枚举

    模拟汇编的操作

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 1e5 + 100;
    vector<int> v;
    int n, m;
    int st[maxn], address, top;
    /*
    void cal(int x){//参数x表示当前选择的数字
        if(v.size()>m||v.size()+(n-x+1)<m)
            return;
        if(x==n+1){
            for (int i = 0; i < v.size();i++){
                printf("%d ", v[i]);
            }
            printf("
    ");
            return;
        }
        //选x
        v.push_back(x);
        cal(x + 1);
        //不选x
        v.pop_back();
        cal(x + 1);
        return;
    }*/
    
    void cal(int x,int ret_addr){//新指令(数字)进栈
        int old_top = top;
        st[++top] = x;
        st[++top] = ret_addr;
        st[++top] = old_top;
    }
    int ret(){
        int ret_addr = st[top - 1];
        top = st[top];//让最后一条语句出栈
        return ret_addr;
    }
    int main(){
        cin >> n >> m;
        cal(1, 0);//数字1进栈
        while(top){
            int x = st[top - 2]; //获取的是数字x
            switch(address){
                case 0:
                    if(v.size()>m||v.size()+(n-x+1)<m){
                        address = ret(); //也就是return;
                        continue;
                    }
                    if(x==n+1){
                        for (int i = 0; i < v.size();i++){
                            printf("%d ", v[i]);
                        }
                        printf("
    ");
                        address = ret();
                        continue;
                    }
                    v.push_back(x);
                    cal(x + 1, 1);
                    address = 0;
                    continue;
                case 1:
                    v.pop_back();
                    cal(x + 1, 2);//返回后从cas2继续执行
                    address = 0;
                    continue;
                case 2:
                    address = ret();
            }
        }
    }
    

    其他

    语法

    断言

    int a = 0;
    assert(a==0);
    程序正常运行
        
    assert(a==1);
    会报出错误,也就是说我所断言的a==1是错误的
    

    最大子段和

    int best=0,sum=0;
    for(int i=0;i<n;i++)
    {
        sum=max(array[i],sum+array[k]);
        best=max(sun,best);
    }
    cout<<best<<endl;
    

    双指针问题

    [CF1304D]Shortest and Longest LIS

    https://codeforces.com/contest/1304/problem/D

    构造最长LIS时,要从最小的数开始填

    构造最长LIS时,要从最大的数开始填

    #include<bits/stdc++.h>
    using namespace std;
    
    const int maxn = 2e5 + 100;
    int ans[maxn];
    int main() {
        int T;
        scanf("%d", &T);
        while(T--) {
            int n;string s;
            cin >> n >> s;
            int num = n, last = 0;
            for (int i = 0; i < n;i++) {
                if(i==n-1||s[i]=='>') {
                    for (int j = i; j >= last;j--) {
                        ans[j] = num--;
                    }
                    last = i + 1;
                }
            }
            for (int i = 0; i < n;i++) {
                printf("%d ", ans[i]);
            }
            printf("
    ");
            num = 1, last = 0;
            for(int i=0;i<n;i++) {
                if(i==n-1||s[i]=='<') {
                    for (int j = i; j >= last;j--) {
                        ans[j] = num++;
                    }
                    last = i + 1;
                }
            }
            for (int i = 0; i < n;i++) {
                printf("%d ", ans[i]);
            }
            printf("
    ");
        }
    }
    

    [CF1199C] mp3

    这题神烦

    https://codeforces.com/contest/1199/problem/C

    #include<bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int maxn = 4e5 + 100;
    vector<int> v;
    map<int, int> mp;
    int pre[maxn];
    int main() {
        int n, k;
        scanf("%d%d", &n, &k);
        k *= 8;
        int m = 0;
        for (int i = 1; i <= n;i++) {
            int x;
            scanf("%d", &x);
            if(!mp[x]) {
                v.push_back(x);
            }
            mp[x]++;
        }
        sort(v.begin(), v.end());
        m = unique(v.begin(), v.end())-v.begin();
        int tot = k / n;
        if(tot>20)
            tot = 20;
        tot = 1 << tot;
        for (int i = 0; i < m;i++){
            if(i==0)
                pre[i] = mp[v[i]];
            else
                pre[i] = pre[i - 1] + mp[v[i]];
        }
        int ans = 0x3f3f3f3f;
        for (int i = 0; i < m;i++) {
            int j = i + tot - 1;
            int x = 0;
            if(i!=0)
                x = pre[i - 1];
            if(j>=m) {
                ans = min(ans, x);
                break;
            }
            int y = pre[j];
            ans = min(ans, pre[m - 1] - y + x);
        }
        ans = max(ans, 0);
        cout<<ans<<endl;
    }
    
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  • 原文地址:https://www.cnblogs.com/guaguastandup/p/12585284.html
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