src/11/6/P14358.cpp

@@ -1,42 +0,0 @@
-#include <algorithm>
-#include <cstdint>
-#include <cstdio>
-#include <functional>
-#include <iostream>
-#include <istream>
-using ll = int64_t;
-#define printf
-
-const ll maxn = 17;
-ll isd=1;
-ll n,m,x=1,y=1,ansx,ansy,ansn,a[maxn*maxn];
-
-int main(){
-    std::iostream::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-
-    std::cin>>n>>m;
-    for(ll i=1;i<=n*m;i++){
-        std::cin>>a[i];
-    }
-    ansn=a[1];
-    std::sort(a+1,a+1+n*m,std::greater<>());
-    ll tot=1;
-    while(y<=m){
-        printf("x=%lld, y=%lld\n",x,y);
-        if(x==n+1||x==0){
-            y++;
-            isd=isd*-1;
-            x+=isd;
-            continue;
-        }
-        ll tmp=a[tot++];
-        printf("tmp=%lld\n",tmp);
-        if(tmp==ansn){
-            ansx=x,ansy=y,ansn=tmp;
-            break;
-        }
-        x+=isd;
-    }
-    std::cout<<ansy<<" "<<ansx<<"\n";
-}

src/11/7/P14359.cpp

@@ -1,40 +0,0 @@
-#include <cstdint>
-#include <cstdio>
-#include <iostream>
-#include <istream>
-
-using ll = int64_t;
-#define printf
-const ll p220=1048576,maxa=p220+7;
-ll n,k,f[maxa],now,tmp,ans,last;
-struct S{
-    ll l,r;
-};
-
-int main(){
-    std::iostream::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-
-    for(ll i=0;i<maxa;i++){
-        f[i]=-1;
-    }
-    std::cin>>n>>k;
-    f[0]=1;
-    last=0;
-    for(ll i=1;i<=n;i++){
-        std::cin>>tmp;
-        now^=tmp;
-        printf("i=%lld, now=%lld\n",i,now);
-        printf("f[%lld]=%lld\n",now^k,f[now^k]);
-        if(f[now^k]!=-1){
-            if(f[now^k]>last){
-                ans++;
-                last=i;
-            }
-            printf("add %lld, %lld\n",f[now^k],i);
-            f[now^k]=-1;
-        }
-        f[now]=i+1;
-    }
-    std::cout<<ans<<"\n";
-}

src/11/7/P14360.cpp

@@ -1,86 +0,0 @@
-#include <iostream>
-#include <vector>
-#include <numeric>
-#include <algorithm>
-
-// 使用标准 long long 类型，并定义常量
-using ll = long long;
-const int MOD = 998244353;
-const int MAXN = 5005;
-
-// 预计算2的幂，用于快速得到子集总数
-ll p2[MAXN];
-
-void precompute_powers(int n) {
-    p2[0] = 1;
-    for (int i = 1; i <= n; ++i) {
-        p2[i] = (p2[i - 1] * 2) % MOD;
-    }
-}
-
-int main() {
-    // OI/ACM 风格的快速输入输出
-    std::ios_base::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-
-    int n;
-    std::cin >> n;
-
-    std::vector<int> a(n);
-    int total_sum = 0;
-    for (int i = 0; i < n; ++i) {
-        std::cin >> a[i];
-        total_sum += a[i];
-    }
-
-    // 1. 排序：这是算法的关键前提
-    std::sort(a.begin(), a.end());
-    
-    // 预计算2的幂
-    precompute_powers(n);
-
-    // 2. DP 数组设置
-    // dp[j] 表示从当前已处理的木棍中选出若干根，使其和为 j 的方案数。
-    // 背包容量上界是所有木棍的总和。
-    std::vector<ll> dp(total_sum + 1, 0);
-    dp[0] = 1; // 和为0的方案只有一种：不选任何木棍
-
-    ll totans = 0;
-    int cursum = 0; // 记录当前处理过的木棍的总和，作为DP容量的上界
-
-    // 3. 主循环：依次处理每根木棍
-    for (int i = 0; i < n; ++i) {
-        // a[i] 是当前木棍的长度
-        
-        // 3.1 统计答案：
-        // 假设 a[i] 是最长边，从前面的 a[0]...a[i-1] 中选边。
-        // 此时 dp 数组存储的是从 {a[0], ..., a[i-1]} 中选择的方案。
-        // 我们需要找到满足 sum(S) > a[i] 的子集 S 的数量。
-        
-        // 计算 sum(S) <= a[i] 的方案数
-        ll nans = 0;
-        // 注意 j 的上界是 cursum 和 a[i] 的较小值
-        for (int j = 0; j <= std::min(cursum, a[i]); ++j) {
-            nans = (nans + dp[j]) % MOD;
-        }
-
-        // 从 {a[0], ..., a[i-1]} 中选子集，总方案数是 2^i
-        ll totpre = p2[i];
-        
-        // 合法方案数 = 总方案数 - 不合法方案数 (sum(S) <= a[i])
-        ll ok = (totpre - nans + MOD) % MOD;
-        
-        // 累加到最终答案
-        totans = (totans + ok) % MOD;
-        
-        // 3.2 更新DP数组
-        // 将 a[i] 加入可选木棍池，为后续 a[i+1], a[i+2]... 作为最长边时做准备。
-        cursum += a[i];
-        // 0/1 背包，必须逆序遍历以保证每个物品只被选一次
-        for (int j = cursum; j >= a[i]; --j) {
-            dp[j] = (dp[j] + dp[j - a[i]]) % MOD;
-        }
-    }
-
-    std::cout << totans << std::endl;
-}

src/11/8/P11362.cpp

@@ -1,59 +0,0 @@
-#include <algorithm>
-#include <cstdint>
-#include <iostream>
-#include <istream>
-#include <set>
-#include <unordered_set>
-using ll = int64_t;
-#define sl static inline
-
-const ll maxm=1e5+7,p=1e9+7;
-ll n,m,v;
-struct X{
-    ll c,d;
-    inline bool operator<(const X& o)const{
-        return c<o.c;
-    }
-}x[maxm];
-std::unordered_set<int> s;
-
-sl void solve(){
-    std::cin>>n>>m>>v;
-    for(ll i=1;i<=m;i++){
-        std::cin>>x[i].c>>x[i].d;
-        s.insert(x[i].c);
-    }
-    std::sort(x+1,x+1+m);
-    for(ll i=1;i<m;i++){
-        if(x[i].c==x[i+1].c && x[i].d!=x[i+1].d){
-            std::cout<<"0\n";
-            return;
-        }
-    }
-    ll ans=1;
-    for(ll i=1;i<n;i++){
-        if(s.count(i)&s.count(i+1)){
-            ans=(ans+1)%p;
-        }else{
-            ans=ans*v%p*v%p;
-        }
-    }
-    std::cout<<ans<<"\n";
-}
-
-/*
-xi yes xi+1 no v*v
-xi no xi+1 yes v*v
-xi yes xi+1 yes v*v-1
-*/
-
-int main(){
-    std::iostream::sync_with_stdio(false);
-    std::cin.tie(nullptr);
-
-    ll t;
-    std::cin>>t;
-    while(t--){
-        solve();
-    }
-}

src/8/11/P11362.cpp

@@ -0,0 +1,84 @@
+#include <iostream>
+#include <map>
+
+using ll = long long;
+using namespace std;
+const int MOD = 1e9 + 7;
+
+ll p(ll b, ll e) {
+    ll res = 1;
+    b %= MOD;
+    while (e > 0) {
+        if (e % 2 == 1) res = (res * b) % MOD;
+        b = (b * b) % MOD;
+        e /= 2;
+    }
+    return res;
+}
+
+void solve() {
+    ll n;
+    int m;
+    ll v;
+    cin >> n >> m >> v;
+    map<int, int> mc;
+    for (int i = 0; i < m; ++i) {
+        int c, d;
+        cin >> c >> d;
+        if (mc.count(c) && mc[c] != d) {
+            for (int j = i + 1; j < m; ++j) cin >> c >> d;
+            cout << 0 << endl;
+            return;
+        }
+        mc[c] = d;
+    }
+    if (m == 0) {
+        if (n == 1) { 
+            cout << 1 << endl;
+        } else {
+            cout << p(v, (n - 1) * 2) << endl;
+        }
+        return;
+    }
+    ll tot = 1;
+    ll vmod = v % MOD;
+    int lp = 0;
+    for (auto const& [idx, val] : mc) {
+        ll len = idx - lp;
+        if (len > 0) {
+            ll s;
+            if (lp == 0) {
+                s = p(vmod, (len - 1) * 2);
+            } else {
+                ll v2len = p(vmod, len * 2);
+                ll vlen = p(vmod, len);
+                ll vlenm1 = p(vmod, len - 1);
+                s = (v2len - vlen + vlenm1 + MOD) % MOD;
+            }
+            tot = (tot * s) % MOD;
+        }
+        lp = idx;
+    }
+
+    
+    if (lp < n) {
+        ll len = n - lp;
+        
+        
+        ll s = p(vmod, len * 2);
+        tot = (tot * s) % MOD;
+    }
+
+    cout << tot << endl;
+}
+
+int main() {
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+    cout.tie(nullptr);
+    int t;
+    cin >> t;
+    while (t--) {
+        solve();
+    }
+}