算法一、DFS一号

using namespace std;
int n = 2, a[5], s;
int dfs(int x, int sum) {
    if (x > n) return sum;
    int i = dfs(x + 1, sum);
    int j = dfs(x + 1, sum + a[x]);
    if (i == s) return i;
    if (j == s) return j;
    return -1;
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    cout << dfs(1, 0) << endl;
    return 0;
}

算法二、DFS二号

using namespace std;
int a, b;
int dfs(int x) {
    if (x <= 5) return x;
    return dfs(x / 2) + dfs(x - x / 2);
} 
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", dfs(a) + dfs(b));
    return 0;
}

算法三、BFS

using namespace std;
int n = 2, a[5], s;
queue<int> q;
void bfs() {
    q.push(0);
    int c = 0;
    while (q.size()) {
        c++;
        int f = q.front(); q.pop();
        if (f == s) {printf("%d\n", f); exit(0);}
        q.push(f + a[c]);
        q.push(f);
    }
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    bfs();
    return 0;
}

算法四、直接算咯

using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

算法五、二分

using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    int l = 0, r = 200000000;
    while (l < r) {
        int mid = l + r >> 1;
        if (mid == a + b) {printf("%d\n", mid); return 0;}
        if (mid <  a + b) l = mid + 1;
        if (mid >  a + b) r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

算法六、稍微有点暴力的枚举

#include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    for (int i = 0;i <= 200000000; i++) if (a + b == i) {printf("%d\n", i); break;}
    return 0;
}

算法七、最短路之dijkstra

#include <bits/stdc++.h>
using namespace std;
int w[5][5], d[5], v[5];
int n = 3;
void dijkstra() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            d[y] = min(d[y], d[x] + w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    dijkstra();
    printf("%d\n", d[3]);
    return 0;
}

算法八、最短路之SPFA

#include <bits/stdc++.h> using namespace std; int a, b, n = 3; int w[5][5], d[5], v[5]; queue q; void spfa() { memset(d, 0x3f, sizeof d); memset(v, 0, sizeof v); d[1] = 0, v[1] = 1; q.push(1); while (q.size()) { int x = q.front(); q.pop(); v[x] = 0; for (int i = 1;i <= n; i++) { // if (w[x][i] == 0x3f) continue; if (d[i] > d[x] + w[x][i]) { d[i] = d[x] + w[x][i]; if (!v[i]) q.push(i), v[i] = 1; } } } } int main() { scanf("%d%d", &a, &b); memset(w, 0x3f, sizeof w); w[1][2] = a; w[2][3] = b; spfa(); printf("%d\n", d[3]); return 0; }

算法九、最短路之Floyd

using namespace std;
int d[5][5], n = 3;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(d, 0x3f, sizeof d);
    d[1][2] = a; d[2][3] = b;
    for (int k = 1;k <= n; k++)
        for (int i = 1;i <= n; i++)
            for (int j = 1;j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    printf("%d\n", d[1][3]);
    return 0;
}

算法十、高精

#include<bits/stdc++.h> using namespace std; string a0, b0; int a[1005], b[1005]; int main(){ cin >> a0 >> b0; int l1 = a0.size(), l2 = b0.size(); for (int i = 0;i < l1; i++) a[l1 - i] = a0[i] - 48; for (int i = 0;i < l2; i++) b[l2 - i] = b0[i] - 48; l1 = max(l1, l2); for (int i = 1;i <= l1; i++) { a[i] += b[i]; if (a[i] > 9) a[i + 1] += 1, a[i] %= 10; } if (a[max(l1, l2) + 1] > 0) l1++; for (int i = l1;i >= 1; i--) printf("%d", a[i]); return 0; }

算法十一、最小生成树之kruskal

using namespace std;
struct rec {
    int x, y, z;
} edge[5];
 
int fa[5], m = 2, ans = 0;
 
int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
int cmp(rec a, rec b) { return a.z < b.z; }
 
int main() {
    int a, b; scanf("%d%d", &a, &b);
    edge[1] = (rec){1, 2, a};
    edge[2] = (rec){2, 3, b};
    for (int i = 1;i <= m + 1; i++) fa[i] = i;
    sort(edge + 1, edge + 1 + m, cmp);
    for (int i = 1;i <= m; i++) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if (x == y) continue;
        fa[x] = y;
        ans += edge[i].z;
    }
    printf("%d\n", ans);
    return 0;
}

算法十二、最小生成树之prim

using namespace std;
int w[5][5], d[5], n = 3, ans, v[5];
 
void prim() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            if (!v[y]) d[y] = min(d[y], w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    prim();
    int ans = 0;
    for (int i = 2;i <= n; i++) ans += d[i];
    printf("%d\n", ans);
    return 0;
}

算法十三、前缀和

using namespace std;
int a[5], s[5];
int main() {
    for (int i = 1;i <= 2; i++) scanf("%d", &a[i]), s[i] += a[i] + s[i - 1];
    printf("%d\n", s[2]);
    return 0;
}

算法十四、后缀和

using namespace std;
int a[5], s[5];
int main() {
    for (int i = 2;i >= 1; i--) scanf("%d", &a[i]), s[i] += a[i] + s[i + 1];
    printf("%d\n", s[1]);
    return 0;
}

算法十五、位运算

using namespace std;
int add(int a, int b) {
    if (b == 0) return a;
    return add(a ^ b, (a & b) << 1);
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d\n", add(a, b));
    return 0;
}

算法十六、树的直径——BFS

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
 
int head[maxn * 2],edge[maxn * 2],Next[maxn * 2],ver[maxn * 2];
int vis[maxn], dist[maxn];
int n = 3, p, q, d;
int tot = 0;
int maxd = 0;
 
void add(int u,int v,int w) {
    ver[ ++ tot] = v,edge[tot] = w;
    Next[tot] = head[u],head[u] = tot;
}
 
int BFS(int u) {
    queue<int>Q;
    while(!Q.empty()) Q.pop();
    memset(vis, 0, sizeof vis);
    memset(dist, 0, sizeof dist);  
    Q.push(u);
    int x, max_num = 0;
    while(!Q.empty()) {
        x = Q.front();
        Q.pop();
        vis[x] = 1;
        for(int i = head[x]; i ; i = Next[i]) {
            int y = ver[i];
            if(vis[y]) continue;
            vis[y] = 1;
            dist[y] = dist[x] + edge[i];
            if(dist[y] > maxd) {
                maxd = dist[y];
                max_num = y;
            }
            Q.push(y);
        }
    }
    return max_num;
}
int main(void) {
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    BFS(BFS(1));
    int ans = 0;
    for (int i = 1;i <= n; i++) ans = max(ans, dist[i]);
    printf("%d\n", ans);
    return 0;
}

算法十七、树的直径——DFS

#include<bits/stdc++.h>
#define maxn 100000
using namespace std;
struct node {
    int u, v, w, nex;
} edge[2 * maxn + 10];
int n = 3, d[maxn + 10], head[maxn + 10], f_num, cnt = 0, ans;
inline void add(int x,int y,int z) {
    edge[++cnt].u = x;
    edge[cnt].v = y;
    edge[cnt].w = z;
    edge[cnt].nex = head[x];
    head[x] = cnt;
}
inline void dfs(int x, int fa) {
    if(ans < d[x]) {
        ans = d[x];
        f_num = x;
    }
    for (int i = head[x]; i != -1; i = edge[i].nex) {
        int j = edge[i].v;
        if (j == fa)continue;
        d[j] = d[x] + edge[i].w;    
        dfs(j, x);
    }
}
int main() {
    memset(head, -1, sizeof(head));
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dfs(1, 0);
    ans = 0;
    d[f_num] = 0;
    dfs(f_num, 0);
    for (int i = 1;i <= n; i++) ans = max(ans, d[i]);
    printf("%d", ans);
    return 0;
}

算法十八、树的直径——树形DP

#include <bits/stdc++.h>
using namespace std;
int f[5], n = 3, cnt, h[5], ans, dis[5];
struct edge {
    int to, next, vi;
} e[5];
void add(int u, int v, int w) {
    e[cnt].to= v;
    e[cnt].vi = w;
    e[cnt].next = h[u];
    h[u] = cnt++;
}
void dp(int u, int fa) {
    for (int i = h[u]; ~i; i = e[i].next) {
        int v = e[i].to;
        if (v == fa) continue;
        dp(v, u);
        ans = max(ans, dis[v] + dis[u] + e[i].vi);
        dis[u] = max(dis[u], dis[v] + e[i].vi);
    }
}
int main() {
    memset(h, -1, sizeof h);
    int a, b; scanf("%d%d", &a, &b);
    add(1, 2, a); add(2, 1, a);
    add(2, 3, b); add(3, 2, b);
    dp(1, 0);
    printf("%d\n", ans);
    return 0;
}

算法十九、网络流

using namespace std;
#define set(x) Set(x)
#define REP(i,j,k) for (int i=(j),_end_=(k);i<=_end_;++i)
#define DREP(i,j,k) for (int i=(j),_start_=(k);i>=_start_;--i)
#define mp make_pair
#define x first
#define y second
#define pb push_back
template<typename T> inline bool chkmin(T &a,const T &b){ return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a,const T &b){ return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef pair<int,int> node;
const int dmax = 1010, oo = 0x3f3f3f3f;
int n, m;
int a[dmax][dmax] , ans;
int d[dmax], e[dmax];
priority_queue <node> q;
inline bool operator >(node a,node b){ return a.y>b.y; }
bool p[dmax];
void Set(int x){ p[x] = 1; }
void unset(int x){ p[x] = 0; }
bool check(int x){ return x != 1 && x != n && !p[x] && e[x] > 0; }
void preflow(){
    e[1] = oo;
    d[1] = n - 1;
    q.push(mp(1, n - 1));
    set(1);
    while (!q.empty()) {
        bool flag = 1;
        int k = q.top().x;
        q.pop(), unset(k);
        DREP(i, n, 1)
        if ((d[k] == d[i] + 1 || k == 1) && a[k][i] > 0){
            flag = 0;
            int t = min(a[k][i], e[k]);
            e[k] -= t;
            a[k][i] -= t;
            e[i] += t;
            a[i][k] += t;
            if (check(i)) {
                q.push(mp(i, d[i]));
                set(i);
            }
            if (e[k] == 0) break;
        }
        if (flag) {
            d[k] = oo;
            REP(i, 1, n)
            if (a[k][i] > 0) chkmin(d[k], d[i] + 1);
        }
        if (check(k)) {
            q.push(mp(k, d[k]));
            set(k);
        }
    }
    ans = e[n];
}
int main() {
    n = 2, m = 2;
    int x, y;
    scanf("%d%d", &x, &y);
    a[1][2] += x + y;
    preflow();
    printf("%d\n", ans);
    return 0;
}

算法二十、线段树

#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
using namespace std;
struct SegmentTree {
    int l, r; //区间左右端点 
    long long sum, add; //sum 区间和  add 延迟标记 
} tree[400010];
int a[100010], n = 1, m = 2;
void build (int p, int l, int r) {
    l(p) = l, r(p) = r;
    if(l == r) {sum(p) = a[l]; return;}
    int mid = l + r >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
void spread(int p) {
    if(add(p)) { //节点p有标记 
        sum(p * 2) += add(p) * (r(p * 2) - l(p * 2) + 1); //更新左子节点信息 
        sum(p * 2 + 1) += add(p) * (r(p * 2 + 1) - l(p * 2 + 1) + 1); //更新右子节点
        add(p * 2) += add(p); //给左子节点打延迟标记 
        add(p * 2 + 1) += add(p); //给右子节点打延迟标记 
        add(p) = 0; //清除p的标记 
    }
}
void change(int p, int l, int r, int d) {
    if(l <= l(p) && r >= r(p)) { //完全覆盖 
        sum(p) += (long long) d * (r(p) - l(p) + 1); //更新节点信息 
        add(p) += d; //给节点打延迟标记 
        return;
    }
    spread(p); //下传延迟标记 
    int mid = l(p) + r(p) >> 1;
    if(l <= mid) change(p * 2, l, r, d);
    if(r > mid) change(p * 2 + 1, l, r, d);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
long long ask(int p, int l, int r) {
    if(l <= l(p) && r >= r(p)) return sum(p);
    spread(p);
    int mid = l(p) + r(p) >> 1;
    long long val = 0;
    if(l <= mid) val += ask(p * 2, l, r);
    if(r > mid) val += ask(p * 2 + 1, l, r);
    return val;
}
int main() {
    a[1] = 0;
    build(1, 1, n);
    while(m--) { 
        int d = 0;
        scanf("%d", &d);
        change(1, 1, 1, d);
    }
    printf("%lld\n", ask(1, 1, 1));
    return 0;
}

还有60多种，写不下了