标签:cst ring nal 选择 mod algo time define ace
构造。
前$n-k$个都是$1$,最后$k$个进行构造,首先选择填与上一个数字一样,如果不可行,那么这一格的值$+1$。
#include<map> #include<set> #include<ctime> #include<cmath> #include<queue> #include<string> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<functional> using namespace std; #define ms(x,y) memset(x,y,sizeof(x)) #define rep(i,j,k) for(int i=j;i<=k;i++) #define per(i,j,k) for(int i=j;i>=k;i--) #define loop(i,j,k) for (int i=j;i!=-1;i=k[i]) #define inone(x) scanf("%d",&x) #define intwo(x,y) scanf("%d%d",&x,&y) #define inthr(x,y,z) scanf("%d%d%d",&x,&y,&z) #define infou(x,y,z,p) scanf("%d%d%d%d",&x,&y,&z,&p) #define lson x<<1,l,mid #define rson x<<1|1,mid+1,r #define mp(i,j) make_pair(i,j) #define ft first #define sd second typedef long long LL; typedef pair<int, int> pii; const int low(int x) { return x&-x; } const int INF = 0x7FFFFFFF; const int mod = 1e9 + 7; const int N = 1e6 + 10; const int M = 1e4 + 1; const double eps = 1e-10; int T, n ,m, k, p; int c[100010]; int a[100010]; long long s; void update(int x,int val) { while(x<=100000) { c[x]=c[x]+val; x=x+low(x); } } int sum(int x) { int res=0; while(x>0) { res=res+c[x]; x=x-low(x); } return res; } int main() { while(~scanf("%d%d",&n,&k)) { scanf("%d",&p); s=0; for(int i=1;i<=n-k;i++) a[i]=1; int now=2,x=0,cnt=n-k; for(int i=n-k+1;i<=n;i++) { a[i] = now; if(cnt*100>=p*(i-1)) { x++; } else { now++; cnt+=x; a[i]=now; x=1; } } for(int i=1;i<=n;i++) s=s+(long long) a[i]; printf("%lld\n",s); for(int i=1;i<=n;i++) { printf("%d",a[i]); if(i<n) printf(" "); else printf("\n"); } } return 0; }
标签:cst ring nal 选择 mod algo time define ace
原文地址:http://www.cnblogs.com/zufezzt/p/6696269.html