【AC】
1 #include <stdio.h> 2 #include <iostream> 3 #include <string.h> 4 #include <algorithm> 5 #include <math.h> 6 using namespace std; 7 typedef long long ll; 8 const double PI = acos(-1.0); 9 struct complex 10 { 11 double r,i; 12 complex(double _r = 0,double _i = 0) 13 { 14 r = _r; i = _i; 15 } 16 complex operator +(const complex &b) 17 { 18 return complex(r+b.r,i+b.i); 19 } 20 complex operator -(const complex &b) 21 { 22 return complex(r-b.r,i-b.i); 23 } 24 complex operator *(const complex &b) 25 { 26 return complex(r*b.r-i*b.i,r*b.i+i*b.r); 27 } 28 }; 29 void change(complex y[],int len) 30 { 31 int i,j,k; 32 for(i = 1, j = len/2;i < len-1;i++) 33 { 34 if(i < j)swap(y[i],y[j]); 35 k = len/2; 36 while( j >= k) 37 { 38 j -= k; 39 k /= 2; 40 } 41 if(j < k)j += k; 42 } 43 } 44 void fft(complex y[],int len,int on) 45 { 46 change(y,len); 47 for(int h = 2;h <= len;h <<= 1) 48 { 49 complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h)); 50 for(int j = 0;j < len;j += h) 51 { 52 complex w(1,0); 53 for(int k = j;k < j+h/2;k++) 54 { 55 complex u = y[k]; 56 complex t = w*y[k+h/2]; 57 y[k] = u+t; 58 y[k+h/2] = u-t; 59 w = w*wn; 60 } 61 } 62 } 63 if(on == -1) 64 for(int i = 0;i < len;i++) 65 y[i].r /= len; 66 } 67 68 const int MAXN = 400040; 69 complex x1[MAXN]; 70 int a[MAXN/4]; 71 long long num[MAXN];//100000*100000会超int 72 char str[MAXN/4]; 73 int main() 74 { 75 freopen("duel.in","r",stdin); 76 freopen("duel.out","w",stdout); 77 int n=0; 78 scanf("%s",str+1); 79 int ls=strlen(str+1); 80 for(int i=1;i<=ls;i++) 81 { 82 if(str[i]=='1') 83 { 84 a[n++]=i; 85 } 86 } 87 memset(num,0,sizeof(num)); 88 for(int i = 0;i < n;i++) 89 { 90 num[a[i]]++; 91 } 92 sort(a,a+n); 93 int len1 = a[n-1]+1; 94 int len = 1; 95 while( len < 2*len1 )len <<= 1; 96 for(int i = 0;i < len1;i++) 97 x1[i] = complex(num[i],0); 98 for(int i = len1;i < len;i++) 99 x1[i] = complex(0,0); 100 fft(x1,len,1); 101 for(int i = 0;i < len;i++) 102 x1[i] = x1[i]*x1[i]; 103 fft(x1,len,-1); 104 for(int i = 0;i < len;i++) 105 num[i] = (long long)(x1[i].r+0.5); 106 len = 2*a[n-1]; 107 //减掉取两个相同的组合 108 for(int i = 0;i < n;i++) 109 num[a[i]+a[i]]--; 110 //选择的无序，除以2 111 for(int i = 1;i <= len;i++) 112 { 113 num[i]/=2; 114 } 115 ll ans=0; 116 for(int i=1;i<=len;i++) 117 { 118 if(i%2!=0) continue; 119 if(str[i/2]=='0') continue; 120 ans+=num[i]; 121 } 122 cout<<ans<<endl; 123 return 0; 124 }
http://www.cnblogs.com/kuangbin/archive/2013/07/24/3210565.html