A. Vasya and Socks http://codeforces.com/contest/460/problem/A
水题暴力
1 #include<cstdio> 2 int main(){ 3 int n,m; 4 while(~scanf("%d%d",&n,&m)){ 5 int ans=0; 6 while(n){ 7 n--; 8 ans++; 9 if(!(ans%m)) n++; 10 } 11 printf("%d ",ans); 12 } 13 return 0; 14 }
B. Little Dima and Equation http://codeforces.com/contest/460/problem/B
sx=x所有位的和，要知道sx在1~1e9的x 对应的是1~81，枚举81，判断。
1 #include<cstdio> 2 typedef __int64 LL; 3 LL gxpow(int x,int n){ 4 LL res=1; 5 while(n--) res*=x; 6 return res; 7 } 8 int sum(LL x){ 9 int res=0; 10 while(x){ 11 res+=x%10; 12 x/=10; 13 } 14 return res; 15 } 16 LL ans[128]; 17 int main(){ 18 int a,b,c; 19 while(~scanf("%d%d%d",&a,&b,&c)){ 20 int la=0; 21 for(int i=1;i<=81;i++){ 22 LL x=b*gxpow(i,a)+c; 23 if(0<x&&x<1e9&&sum(x)==i){ 24 ans[la++]=x; 25 } 26 } 27 printf("%d ",la); 28 if(!la) continue; 29 for(int i=0;i<la;i++){ 30 printf("%d ",ans[i]); 31 } 32 puts(""); 33 } 34 return 0; 35 }
C. Present http://codeforces.com/contest/460/problem/C
二分答案，注意LR上下界，瞎定义tle了一次，对答案贪心的从前往后浇水，看能否构成答案，区间更新on左++右--是线段树nlogn的10分之一的时间46ms vs 390ms。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #define mt(a,b) memset(a,b,sizeof(a)) 5 using namespace std; 6 const int M=100010; 7 int n,m,w,a[M],lazy[M]; 8 bool judge(int ans){ 9 mt(lazy,0); 10 int add=0,left=m; 11 for(int i=1;i<=n;i++){ 12 add+=lazy[i]; 13 int now=a[i]+add; 14 if(now<ans){ 15 int need=ans-now; 16 if(need>left) return false; 17 left-=need; 18 if(w>1){ 19 lazy[i+1]+=need; 20 if(i+w<=n) 21 lazy[i+w]-=need; 22 } 23 } 24 } 25 return true; 26 } 27 int main(){ 28 while(~scanf("%d%d%d",&n,&m,&w)){ 29 int L=1,R=0; 30 for(int i=1;i<=n;i++){ 31 scanf("%d",&a[i]); 32 R=max(R,a[i]); 33 } 34 R+=m; 35 while(L<=R){ 36 int mid=(L+R)>>1; 37 if(judge(mid)) 38 L=mid+1; 39 else 40 R=mid-1; 41 } 42 printf("%d ",L-1); 43 } 44 return 0; 45 }
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