标签：unix sed ring 展示 否则 cstring png mes ios
2018/8/15 qbxt 测试
期望得分：100；实际得分：50 不知道为什么写挂了，明明是个水题 T^T
思路：模拟
注意：如果用 char 类型存储的话，如果有‘z‘ + 9 会爆char 但是我明明用的 string 啊
#include<algorithm> #include<iostream> #include<cstring> #include<cstdio> using namespace std; int n, m; int len; string s, ss; int main() { scanf("%d%d", &n, &m); cin >> s >> ss; len = s.length(); for(int i = 0; i < len; i++) { s[i] += ss[i%m] - ‘0‘; if(s[i] > ‘z‘) s[i] = s[i] - ‘z‘ + ‘a‘ - 1; } cout << s << ‘\n‘; return 0; }
#include <cstdio> #define next(i) ((i) == K-1 ? 0 : (i) + 1) int main() { int L, K; scanf("%d%d", &L, &K); char * s1 = new char[L + 2]; char * s2 = new char[K + 2]; scanf("%s%s", s1, s2); int j = 0; for (int i=0; i<L; i++, j = (j==K-1 ? 0 : j+1)) s1[i] = ((s1[i] - ‘a‘) + (s2[j] - ‘0‘)) % 26 + ‘a‘; puts(s1); return 0; }
期望得分：100；实际得分：100
思路：将因为是完全平方数，所以这个数的因子在1~n中一定是出现了偶数次，道理显然，否则一定不会出现这样一个数。因此我们可以对于N分解质因数，用线性筛O(N)求出1~n的素数。求出素数后用N除以每个素数，开一个数组记录一下出现的次数，对于除出来的商，我们在判断一下能否在继续除以当前的素数，能继续除就继续加，道理显然。
因此，如果数组中记录的数为奇数，就-1，这样就能保证以上偶数次的要求。
#include<iostream> #include<cstdlib> #include<cstdio> #define LL long long using namespace std; const int maxn = 5000005; const LL mod = 100000007; bool not_prime[maxn]; int cnt; LL n, ans=1; LL prime[maxn]; LL num[maxn]; LL ksm(LL a, LL b) { if(b == 0) return 1ll; if(b == 1) return a%mod; LL tmp = ksm(a, b/2)%mod; if(b%2 == 0) return ((tmp%mod)*(tmp%mod))%mod; else return ((((tmp%mod)*tmp)%mod)*(a%mod))%mod; } int main() { scanf("%lld", &n); not_prime[1] = true; for(LL i = 2; i <= n; i++) { if(!not_prime[i]) prime[++cnt] = i; for(int j = 1; j <= cnt; j++) { if(prime[j]*i>n) break; not_prime[prime[j]*i] = true; if(i%prime[j] == 0) break; } } for(int i = 1; i <= cnt; i++) { LL aa = n; while(aa != 0) { num[i] += aa/prime[i]; aa /= prime[i]; } } for(int i = 1; i <= cnt; i++) { if(num[i]%2 == 0) ans = (ans*ksm(prime[i], num[i]))%mod; else ans = (ans*ksm(prime[i], num[i]-1))%mod; } printf("%lld", ans); return 0; }
期望。。。不要爆零 实际。。。8分 qwq
思路：一看就是图论题，然后手动模拟了一下样例1，开始码代码。。发现后两个样例过不了我居然还以为样例错了，还去问老师。。傻的一批
正解：跑两边Floyd，第一次不考虑换马的情况求出最短路，第二次则要考虑换马 f[i][j] = min(f[i][k]+f[k][j]);
#include<algorithm> #include<cstring> #include<cstdio> #include<queue> #define M 10005 #define MAXN 0x3f3f3f using namespace std; queue<int> q; int s, f; int n, m, tot; int e[M], v[M]; double dis[M], cap[M]; int to[M], net[M], head[M], vis[M]; void add(int u, int v, double w) { to[++tot] = v; net[tot] = head[u]; head[u] = tot; cap[tot] = w; } void spfa(int x) { for(int i = 1; i <= n; i++) dis[i] = 1.0*MAXN, vis[i] = 0; dis[x] = 0; vis[x] = 1; q.push(x); while(!q.empty()) { int y = q.front(); q.pop(); vis[y] = 0; for(int i = head[y]; i; i = net[i]) { int t = to[i]; if(dis[t] > dis[y] + cap[i]) { dis[t] = dis[y] + cap[i]; if(!vis[t]) vis[t] = 1, q.push(t); } } } // for(int i = 1; i <= n; i++) printf("%lf\n", dis[i]); } int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; i++) scanf("%d%d", &e[i], &v[i]); for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) { int a; scanf("%d", &a); if(a == -1) continue; else if(e[i] < a) continue; else { double tmp = 1.0 * a / v[i]; add(i, j, tmp); } } for(int i = 1; i <= m; i++) { scanf("%d%d", &s, &f); spfa(s); printf("%.6lf\n", dis[f]); } return 0; }
#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; const int maxn=110; int n,q,e[maxn],s[maxn]; double dis[maxn][maxn],dist[maxn][maxn]; int main() { int T=1; for (int t=1;t<=T;t++) { scanf("%d%d",&n,&q); for (int a=1;a<=n;a++) scanf("%d%d",&e[a],&s[a]); for (int a=1;a<=n;a++) for (int b=1;b<=n;b++) { scanf("%lf",&dis[a][b]); if (dis[a][b]<0) dis[a][b]=1e+20; if (a==b) dis[a][b]=0; } for (int a=1;a<=n;a++) for (int b=1;b<=n;b++) for (int c=1;c<=n;c++) dis[b][c]=min(dis[b][c],dis[b][a]+dis[a][c]); for (int a=1;a<=n;a++) for (int b=1;b<=n;b++) dist[a][b]=1e+20; for (int a=1;a<=n;a++) dist[a][a]=0; for (int a=1;a<=n;a++) for (int b=1;b<=n;b++) if (dis[a][b]<=e[a]) dist[a][b]=dis[a][b]/s[a]; for (int a=1;a<=n;a++) for (int b=1;b<=n;b++) for (int c=1;c<=n;c++) dist[b][c]=min(dist[b][c],dist[b][a]+dist[a][c]); //printf("Case #%d:",t); for (int a=1;a<=q;a++) { int s,e; scanf("%d%d",&s,&e); printf("%.6lf\n",dist[s][e]); } } return 0; }
期望。。。不要爆零 实际。。骗分得到了10分 因为测试点数量比上一个少 qwq
思路：类似于滑动窗口，然后通过树状数组求解
树状数组似撒？能吃吗？qwq
骗分代码我就不展示了 捂脸
#include<cstdio> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #ifdef unix #define LL "%lld" #else #define LL "%I64d" #endif #define lb(x) ((x)&(-(x))) const int maxn=100010; int n,z[maxn],y[maxn],x[maxn]; long long k; bool cmp(int a,int b) { return z[a]<z[b]; } void insert(int *z,int p,int d) { for (; p<=n; p+=lb(p)) z[p]+=d; } int query(int *z,int p) { int ans=0; for (; p; p-=lb(p)) ans+=z[p]; return ans; } int main() { scanf("%d" LL,&n,&k); for (int a=1; a<=n; a++) scanf("%d",&z[a]),y[a]=a; sort(y+1,y+n+1,cmp); x[y[1]]=1; for (int a=2; a<=n; a++) if (z[y[a]]==z[y[a-1]]) x[y[a]]=x[y[a-1]]; else x[y[a]]=x[y[a-1]]+1; for (int a=1; a<=n; a++) z[a]=x[a]; memset(x,0,sizeof(x)); memset(y,0,sizeof(y)); long long nowans=0; int p=n; while (p>=1) { nowans+=query(y,z[p]-1); insert(y,z[p],1); p--; } p++; long long ans=0; for (int a=1; a<=n; a++) { if (p==a) { nowans-=a-1-query(x,z[p])+query(y,z[p]-1); insert(y,z[p],-1); p++; } nowans+=a-1-query(x,z[a])+query(y,z[a]-1); insert(x,z[a],1); while (nowans>k && p<=n) { nowans-=a-query(x,z[p])+query(y,z[p]-1); insert(y,z[p],-1); p++; } if (nowans<=k) ans+=n-p+1; } printf(LL "\n",ans); return 0; }
标签：unix sed ring 展示 否则 cstring png mes ios
原文地址：https://www.cnblogs.com/v-vip/p/9483512.html
