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AddAndMultiply.cpp
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AddAndMultiply.cpp
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#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define FOR(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define SZ(c) ((int)(c).size())
typedef long long LL;
int MOD = 1e9+7;
int p[33][1005], up[33]={};
void add(int &x, LL v) {
x = (x+v)%MOD;
}
const int MXB = 1000;
void pre(int S) {
if(up[S]) return;
up[S] = 1;
p[S][0] = 1;
for(int l=S;l<=MXB;l++) {
for(int j=MXB;j>=l;j--)
add(p[S][j], p[S][j-l]);
}
for(int j=1;j<=MXB;j++) add(p[S][j], p[S][j-1]);
//printf("build: "); for(int j=0;j<=MXB;j++) printf("%d ", p[S][j]); puts("");
}
void solve() {
int A, B, s;
LL t;
scanf("%d%d", &A, &B);
if (A == 0) {
pre(1);
printf("%d\n", (p[1][B]+MOD-1)%MOD);
return;
}
for(t=2,s=1;t<=A;t*=2,s++);
t *= 2; s++;
pre(s);
//printf("s=%d\n", s);
int ans = 0;
LL last = 0, last2 = t/2+A+1;
for(int u=0;u<s && u<=B;u++) {
LL nxt = min(t, (1LL<<u) + A + (u>0));
LL nxt2 = min(t, (1LL<<u) + A + t/2 + (u>0));
add(ans, (nxt-last)*1LL*p[s][B-u]);
if(u+s-1<=B) add(ans, (nxt2-last2)*1LL*p[s][B-(u+s-1)]);
//printf("u=%d: [%I64d, %I64d) %d; [%I64d, %I64d) \n", u, last, nxt, p[s][B-u], last2, nxt2);
last = nxt;
last2 = nxt2;
}
//printf("nxt=%I64d\n", last);
printf("%d\n", (ans+MOD-1)%MOD);
}
int main(void) {
int T;
scanf("%d", &T);
while(T--) solve();
return 0;
}